Using Beta Function and Factorization to Find New π-Series

1. Introduction

The classical rational Ramanujan-type series for $\pi$ introduced by Cooper[6] have the form

$$\sum _{k=0}^{\infty }\frac{bk+c}{m^k}a\left(k\right)=\frac{\lambda \sqrt{d}}{\pi }.$$

In 1905, Glaisher[7] proved that

$$\sum _{k=0}^{\infty }\frac{(4k-1){\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}^4}{{(2k-1)}^4{256}^k}=-\frac{8}{{\pi }^2}.$$

Chan, Chan and Liu[4] proved that

$$\sum _{n=0}^{\infty }\frac{5n+1}{{64}^n}{D}_n=\frac{8}{\sqrt{3}\pi },$$

where $D_n$ denotes the Domb number ${\sum }_{k=0}^n{\left(\genfrac{}{}{0pt}{}{n}{k}\right)}^2\left(\genfrac{}{}{0pt}{}{2k}{k}\right)\left(\genfrac{}{}{0pt}{}{2n-2k}{n-k}\right)$.

Let $b,c\in \mathbb{Z}$, for each $n\in \mathbb{N}$, the coefficient of $x^n$ in the expansion of ${(x^2+bx+c)}^n$ is denoted by $T_n(b,c)$, Sun[11] gave many conjectural series for $\frac{1}{\pi }$ containing the $T_n(b,c)$ such as

$$\sum _{k=0}^{\infty }\frac{3990k+1147}{{(-288)}^{3k}}{T}_k{(62,{95}^2)}^3=\frac{432}{95\pi }(94\sqrt{2}+195\sqrt{14}).$$

In[10], Sun derived several identities involving $\pi$ by the telescoping method. For example, from Bauer’s series[2]

$$\sum _{k=0}^{\infty }(4k+1)\frac{{\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}^3}{{(-64)}^k}=\frac{2}{\pi }$$

and the telescoping sum

$$\sum _{k=0}^n\frac{(16k^3-4k^2-2k+1){\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}^2}{{(2k-1)}^2{(-64)}^k}=\frac{8(2n+1)}{{(-64)}^n}{\left(\genfrac{}{}{0pt}{}{2n}{n}\right)}^3,$$

he deduced

$$\sum _{k=0}^{\infty }\frac{k(4k-1){\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}^3}{{(2k-1)}^2{(-64)}^k}=-\frac{1}{\pi }.$$

(1)

By Gosper’s algorithm[8], Hou and Li[9] give a systematic method to construct $\pi$-series of the form (1). For more results on the $\pi$-series, we refer to [3][5] .

In 1974 Gosper announced the new identity

$$\sum _{k=0}^{\infty }\frac{25k-3}{2^k\left(\genfrac{}{}{0pt}{}{3k}{k}\right)}=\frac{\pi }{2}.$$

Motivated by Gosper’s identity, Almkvist, Krattenthaler and Petersson[1] found some new identities of the type

$$\sum _{k=0}^{\infty }\frac{P\left(k\right)}{\left(\genfrac{}{}{0pt}{}{mk}{nk}\right)x^k}=\pi ,$$

where $P\left(k\right)$ is a polynomial in k. If $P\left(k\right)$ has degree d, they also get that

$$\sum _{k=0}^{\infty }\frac{P\left(k\right)}{\left(\genfrac{}{}{0pt}{}{mk}{nk}\right)x^k}={\int }_0^1\frac{S(t,x)}{{(t^n{(1-t)}^{m-n}-x)}^{d+2}}dt,$$

(2)

where $S(t,x)$ is a polynomial in t of degree $m(d+1)$. They find a good way to get $\pi$ is to have $arctan\left(t\right)$ or $arctan(t-1)$ after integration of (2). Then they get $t^n{(1-t)}^{m-n}-x$ must have the factor $t^2+1$ or ${(t-1)}^2+1$. This restricts m and n and gives the value of x.

Using the method of the beta function, Sun[12] has recently evaluated some series of the type ${\sum }_{k=0}^{\infty }(ak+b)x^k/\left(\genfrac{}{}{0pt}{}{mk}{nk}\right)$. For example, he completely determined the values of

$$\sum _{k=1}^{\infty }\frac{k^r{x}^k}{\left(\genfrac{}{}{0pt}{}{3k}{k}\right)}\left(-\frac{27}{4}<x<\frac{27}{4}\right)$$

for $r=0,\pm 1$.

We first consider the convergence of the left of the equality (2), by Stirling’s formula

$$k!\sim \sqrt{2\pi k}{\left(\frac{k}{e}\right)}^k,k\to +\infty .$$

If $m>n>0$ are integers, then $k\to +\infty$, we have

$$1/\left(\genfrac{}{}{0pt}{}{mk}{nk}\right)\sim \frac{\sqrt{2\pi n(m-n)k}}{\sqrt{m}}\left(\frac{n^n{(m-n)}^{m-n}}{m^m}\right).$$

So we get that if

$$\left|x\right|>\frac{n^n{(m-n)}^{m-n}}{m^m},$$

(3)

the left of the equality (2) is convergent.

The Gamma function is pointed out by Euler as

$$\Gamma \left(x\right)={\int }_0^{+\infty }{t}^{x-1}{e}^{-t}dt,(x>0).$$

The beta function is defined as

$$B(a,b)={\int }_0^1{x}^{a-1}{(1-x)}^{b-1}dx,a>0,b>0.$$

The connection between Gamma function and beta function is given by Euler as

$$B(a,b)=\frac{\Gamma \left(a\right)\Gamma \left(b\right)}{\Gamma (a+b)}.$$

Now we present an auxiliary proposition.

Proposition 1. 

Let $m>n>0$ be integers, $0\le t\le 1$ and $\left|x\right|>n^n{(m-n)}^{m-n}/m^m$, then we have

$$|\frac{t^n{(1-t)}^{m-n}}{x}|<1.$$

Proof. 

Note that for $0\le t\le 1$, we have

$$\sqrt[m]{{\left(\frac{t}{n}\right)}^n{\left(\frac{1-t}{m-n}\right)}^{m-n}}\le \frac{n·\frac{t}{n}+(m-n)·\frac{1-t}{m-n}}{m}=\frac{1}{m},$$

Hence we have

$$|\frac{t^n{(1-t)}^{m-n}}{x}|\le \frac{n^n{(m-n)}^{m-n}}{m^m}\left|\frac{1}{x}\right|<1.$$

By Proposition (1) and beta function we get the following lemma we will use later. □

Lemma 1. 

Let $m>n>0$ be integers, and let x be real numbers with $\left|x\right|>\left(n^n{(m-n)}^{m-n}\right)/m^m$, then we have

$$\sum _{k=0}^{\infty }\frac{1}{x^k\left(\genfrac{}{}{0pt}{}{mk}{nk}\right)}={\int }_0^1\frac{x[(m-1)t^{m-n}{(1-t)}^n+x)]}{{(t^{m-n}{(1-t)}^n-x)}^2}dt.$$

Proof. 

Clearly we have

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }\frac{1}{x^k\left(\genfrac{}{}{0pt}{}{mk}{nk}\right)}& =\sum _{k=0}^{\infty }\frac{\left(nk\right)!\left(\right(m-n\left)k\right)!}{x^k\left(mk\right)!}\hfill \\ \hfill & =\sum _{k=0}^{\infty }\frac{(mk+1)\Gamma (nk+1)\Gamma \left(\right(m-n)k+1)}{x^k\Gamma (mk+2)}\hfill \\ \hfill & =\sum _{k=0}^{\infty }\frac{mk+1}{x^k}B(nk+1,(m-n)k+1)\hfill \\ \hfill & =\sum _{k=0}^{\infty }\frac{mk+1}{x^k}{\int }_0^1{t}^{(m-n)k}{(1-t)}^{nk}dt\hfill \\ \hfill & ={\int }_0^1\sum _{k=0}^{\infty }(mk+1){\left(\frac{t^{m-n}{(1-t)}^n}{x}\right)}^{k}dt.\hfill \end{array}$$

Let

$$y=\frac{t^{m-n}{(1-t)}^n}{x}.$$

Then by Proposition 1 we have $\left|y\right|<1$, then we get

$$\begin{array}{cc}\hfill & {\int }_0^1\sum _{k=0}^{\infty }(mk+1)y^{k}dt\hfill \\ \hfill =& {\int }_0^1\frac{(m-1)y+1}{{(1-y)}^2}dt\hfill \\ \hfill =& {\int }_0^1\frac{x[(m-1)t^{m-n}{(1-t)}^n+x)]}{{(t^{m-n}{(1-t)}^n-x)}^2}dt.\hfill \end{array}$$

□

Using the same method of Lemma 1, we get more equalities we will use later.

Lemma 2. 

Let $m>n>0$ be integers, and let x be real numbers with $\left|x\right|>\left(n^n{(m-n)}^{m-n}\right)/m^m$, then we have

$$\sum _{k=0}^{\infty }\frac{1}{(mk+1)x^k\left(\genfrac{}{}{0pt}{}{mk}{nk}\right)}={\int }_0^1\frac{-x}{t^{m-n}{(1-t)}^n-x}dt.$$

(4)

$$\sum _{k=0}^{\infty }\frac{k}{x^k\left(\genfrac{}{}{0pt}{}{mk}{nk}\right)}={\int }_0^1-\frac{t^{m-n}{(1-t)}^n[(m-1)t^{m-n}{(1-t)}^n+(m+1)x]x}{{(t^{m-n}{(1-t)}^n-x)}^3}dt.$$

(5)

$$\sum _{k=1}^{\infty }\frac{1}{k{x}^k\left(\genfrac{}{}{0pt}{}{mk}{nk}\right)}={\int }_0^1-\frac{n{t}^{m-n}{(1-t)}^{n-1}}{[t^{m-n}{(1-t)}^n-x]}dt.$$

(6)

Proof. 

The proof of equality (4) and (5) is similar as the Lemma 1. To prove the equality (6), we have

$$\begin{array}{cc}\hfill \sum _{k=1}^{\infty }\frac{1}{k{x}^k\left(\genfrac{}{}{0pt}{}{mk}{nk}\right)}& =\sum _{k=1}^{\infty }\frac{1}{k{x}^k}\frac{\left(nk\right)!\left(\right(m-n\left)k\right)!}{\left(mk\right)!}\hfill \\ \hfill & =\sum _{k=1}^{\infty }\frac{nk}{k{x}^k}\frac{\Gamma \left(nk\right)\Gamma \left(\right(m-n)k+1)}{\Gamma (mk+1)}\hfill \\ \hfill & =\sum _{k=1}^{\infty }\frac{n}{x^k}B(nk,(m-n)k+1)\hfill \\ \hfill & =\sum _{k=1}^{\infty }n{\int }_0^1\frac{t^{(m-n)k}{(1-t)}^{nk-1}}{x^k}dt\hfill \\ \hfill & ={\int }_0^1\sum _{k=1}^{\infty }\frac{n}{1-t}{\left(\frac{t^{m-n}{(1-t)}^n}{x}\right)}^{k}dt.\hfill \end{array}$$

Let

$$y=\frac{t^{m-n}{(1-t)}^n}{x}.$$

Then by Proposition 1 we have $\left|y\right|<1$, then we get

$$\begin{array}{cc}\hfill & {\int }_0^1\sum _{k=1}^{\infty }\frac{n}{1-t}{y}^{k}dt\hfill \\ \hfill =& {\int }_0^1\frac{ny}{(1-t)(1-y)}dt\hfill \\ \hfill =& {\int }_0^1-\frac{n{t}^{m-n}{(1-t)}^{n-1}}{[t^{m-n}{(1-t)}^n-x]}dt.\hfill \end{array}$$

By equality (2) and Lemma 1 and 2, we find the denominator of the integration contains the form $t^{m-n}{(1-t)}^n-x$, if we integrate directly, as x is a symbol we just don’t know, in most times we will fail. □

In this paper, we replace x with a suitable rational function $x\left(b\right)$, then by Mathematica we can integrate the rational function $f(t,b)$ from 0 to 1 which the denominator contains the form $t^{m-n}{(1-t)}^n-x\left(b\right)$. Since after integration we get equalities with b and the variable t no longer exists. Then we combine these equalities and let b be a suitable number, this gives lights to find new $\pi$-series. we will show how to get $\pi$-series containing the type $1/\left(\genfrac{}{}{0pt}{}{2k}{k}\right)$, $1/\left(\genfrac{}{}{0pt}{}{3k}{k}\right)$, $1/\left(\genfrac{}{}{0pt}{}{4k}{k}\right)$, $1/\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)$. For example, we get new $\pi$-series as follows

Example 1. 

$$\sum _{k=0}^{\infty }\frac{(-3+2k)2^k}{\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}=\frac{\pi }{2}.$$

$$\sum _{k=1}^{\infty }\frac{(9+35k)8^{k-1}}{k(1+3k)3^k\left(\genfrac{}{}{0pt}{}{3k}{k}\right)}=\frac{\sqrt{3}}{2}\pi -1.$$

$$\sum _{k=1}^{\infty }\frac{(-64+211k+1805k^2)9^k}{k(1+4k)8^k\left(\genfrac{}{}{0pt}{}{4k}{k}\right)}=45+16\sqrt{3}\pi .$$

$$\sum _{k=1}^{\infty }\frac{3+14k}{2k(1+4k){(-3)}^k\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}=\frac{1}{6}(-6+\sqrt{3}\pi ).$$

$$\sum _{k=1}^{\infty }\frac{(1+6k)2^{2k-1}}{k(1+4k)\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}=\frac{\pi }{2}-1.$$

2. $\pi$-series containing

In this section, we will give methods to find $\pi$-series containing the type $1/\left(\genfrac{}{}{0pt}{}{2k}{k}\right)$.

Theorem 1. 

If $b\ne 0$, we have

$$\sum _{k=0}^{\infty }\frac{1}{{(b^2+1/4)}^k\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}=\frac{1+4b^2}{8b^3}\left(2b+arctan\left(\frac{1}{2b}\right)\right).$$

(7)

If $\left|b\right|>\frac{\sqrt{2}}{2}$ , we have

$$\sum _{k=0}^{\infty }\frac{1}{{(1/4-b^2)}^k\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}=\frac{4b^2-1}{8b^3}\left(2b-arctanh\left(\frac{1}{2b}\right)\right),$$

Proof. 

By Lemma 1 we have

$$\sum _{k=0}^{\infty }\frac{1}{x^k\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}={\int }_0^1\frac{-x(t^2-t-x)}{{(t^2-t+x)}^2}dt.$$

By inequality (3) we have $\left|x\right|>\frac{1}{4}$. □

Since $t^2-t+x={(t-\frac{1}{2})}^2+x-\frac{1}{4}$. If $x>\frac{1}{4}$ we let $x-\frac{1}{4}=b^2$, then we $b\ne 0$ and by Mathematica we have

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }\frac{1}{x^k\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}& ={\int }_0^1\frac{-x(t^2-t-x)}{{(t^2-t+x)}^2}dt\hfill \\ \hfill & ={\int }_0^1\frac{-(b^2+\frac{1}{4})(t^2-t-b^2-\frac{1}{4})}{{({(t-\frac{1}{2})}^2+b^2)}^2}dt\hfill \\ \hfill & =\frac{1+4b^2}{8b^3}\left(2b+arctan\left(\frac{1}{2b}\right)\right).\hfill \end{array}$$

If $x<-\frac{1}{4}$ we let $x-\frac{1}{4}=-b^2$, then $\left|b\right|>\frac{\sqrt{2}}{2}$ and by Mathematica we have

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }\frac{1}{x^k\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}& ={\int }_0^1\frac{-x(t^2-t-x)}{{(t^2-t+x)}^2}dt\hfill \\ \hfill & ={\int }_0^1\frac{-(b^2-\frac{1}{4})(t^2-t+b^2+\frac{1}{4})}{{({(t-\frac{1}{2})}^2-b^2)}^2}dt\hfill \\ \hfill & =\frac{4b^2-1}{8b^3}\left(2b-arctanh\left(\frac{1}{2b}\right)\right).\hfill \end{array}$$

Using the same method of Theorem 1 and by Lemma 2 we have

Theorem 2. 

If $b\ne 0$, we have

$$\sum _{k=0}^{\infty }\frac{1}{(2k+1){(b^2+1/4)}^k\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}=\frac{1+4b^2}{2b}arctan\left(\frac{1}{2b}\right).$$

(8)

$$\sum _{k=0}^{\infty }\frac{k}{{(b^2+1/4)}^k\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}=\frac{(1+4b^2)[6b+(3+4b^2)arctan\left(\frac{1}{2b}\right)]}{64b^5}.$$

(9)

$$\sum _{k=1}^{\infty }\frac{1}{k{(b^2+1/4)}^k\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}=\frac{arctan\left(\frac{1}{2b}\right)}{b}.$$

(10)

If $\left|b\right|>\frac{\sqrt{2}}{2}$, we have

$$\sum _{k=0}^{\infty }\frac{1}{(2k+1){(1/4-b^2)}^k\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}=\frac{4b^2-1}{2b}arctanh\left(\frac{1}{2b}\right).$$

$$\sum _{k=0}^{\infty }\frac{k}{{(1/4-b^2)}^k\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}=-\frac{(4b^2-1)[6b+(4b^2-3)arctanh\left(\frac{1}{2b}\right)]}{64b^5}.$$

$$\sum _{k=1}^{\infty }\frac{1}{k{(1/4-b^2)}^k\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}=-\frac{arctanh\left(\frac{1}{2b}\right)}{b}.$$

By Theorem 1 and 2 we can get propositions of equality.

Proposition 2. 

If $b\ne 0$, Via $(-3)\times \left(\right)+8b^2\times \left(\right)$ we have

$$\sum _{k=0}^{\infty }\frac{24b^2-k}{{(1/4+b^2)}^k\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}=\frac{1+4b^2}{2b}arctan\left(\frac{1}{2b}\right)$$

(11)

By equality (8) and (11) we let $b=\frac{1}{2},\frac{\sqrt{3}}{6},\frac{\sqrt{3}}{2}$, we can get series involves $\pi$, here are the examples.

Example 2. 

$$\sum _{k=0}^{\infty }\frac{2^k}{(1+2k)\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}=\frac{\pi }{2}.$$

$$\sum _{k=0}^{\infty }\frac{3^k}{(1+2k)\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}=\frac{4\pi }{3\sqrt{3}}.$$

$$\sum _{k=0}^{\infty }\frac{1}{(1+2k)\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}=\frac{2\pi }{3\sqrt{3}}.$$

$$\sum _{k=0}^{\infty }\frac{(-3+2k)2^k}{\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}=\frac{\pi }{2}.$$

$$\sum _{k=0}^{\infty }\frac{(-9+2k)3^k}{\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}=\frac{4\pi }{\sqrt{3}}.$$

$$\sum _{k=0}^{\infty }\frac{(-1+2k)}{\left(\genfrac{}{}{0pt}{}{2k}{k}\right)}=\frac{2\pi }{9\sqrt{3}}.$$

3. $\mathbf{\pi }$-series containing $\mathbf{1}/\left(\genfrac{}{}{0pt}{}{\mathbf{3}\mathbf{k}}{\mathbf{k}}\right)$

In this section, we will give methods to find $\pi$-series containing the type $1/\left(\genfrac{}{}{0pt}{}{3k}{k}\right)$.

We first give an auxiliary lemma we will use.

Lemma 3. 

Let $f\left(t\right)=t^3-t^2+x$, if $\left|x\right|>\frac{4}{27}$, let $x=-2b{(1+2b)}^2$, then we can factor $f\left(t\right)$ as

$$f\left(t\right)=(t-1-2b)({(t+b)}^2+b(2+3b)),$$

where $b<-\frac{2}{3}$ or $b>\frac{1}{6}\left({(1+\sqrt{2})}^{2/3}+\frac{1}{{(1+\sqrt{2})}^{2/3}}\right)-\frac{1}{3}\approx 0.059$.

Proof. 

Consider $f\left(t\right)=0$, since $\left|x\right|>\frac{4}{27}$, it is easy to calculate that

$$\Delta =\frac{x(27x-4)}{108}>0$$

Then $f\left(t\right)=0$ has one real roots and one pair of unequal conjugate complex roots. Suppose we factor $f\left(t\right)$ as

$$f\left(t\right)=(t+a)({(t+b)}^2+c),$$

then we have

$$\left\{\begin{array}{c}a+2b=-1\hfill \\ 2ab+b^2+c=0\hfill \\ a(b^2+c)=x\hfill \end{array}\right.$$

That is to say $a=-(1+2b)$ and $c=b(3b+2)$ and $x=-2b{(1+2b)}^2$. Since $\left|x\right|>\frac{4}{27}$, we have ${|-2b(1+2b)}^2|>\frac{4}{27}$, then we get $b<-\frac{2}{3}$ or $b>\frac{1}{6}\left({(1+\sqrt{2})}^{2/3}+\frac{1}{{(1+\sqrt{2})}^{2/3}}\right)-\frac{1}{3}$. □

Theorem 3. 

Let μ be defined by

$$\mu =\frac{1}{6}\left({(1+\sqrt{2})}^{2/3}+\frac{1}{{(1+\sqrt{2})}^{2/3}}\right)-\frac{1}{3}.$$

If we define $q\left(b\right)$ as

$$q\left(b\right)=\left\{\begin{array}{cc}arctan\left(\sqrt{\frac{2+3b}{b}}\frac{1}{4b+3}\right),\hfill & b<-\frac{3}{4},orb>\mu \hfill \\ -\frac{\pi }{2},\hfill & b=-\frac{3}{4}\hfill \\ arctan\left(\sqrt{\frac{2+3b}{b}}\frac{1}{4b+3}\right)-\pi ,\hfill & -\frac{3}{4}<b<-\frac{3}{2}\hfill \end{array}\right.$$

If $b>\mu$ or $b<-\frac{2}{3}$, we have

$$\begin{array}{cc}\hfill \sum _{k=0}^{\infty }\frac{1}{\left(\genfrac{}{}{0pt}{}{3k}{k}\right){(-2b{(1+2b)}^2)}^k}=& \frac{27b{(1+2b)}^2}{(2+3b){(1+6b)}^2}+\frac{6b(1+2b)}{{(1+6b)}^3}log\left(\frac{2b}{1+2b}\right)\hfill \\ \hfill +& \frac{2(1+2b)(18b^2+6b-1)}{(2+3b){(1+6b)}^3}\sqrt{\frac{b}{2+3b}}q\left(b\right),\hfill \end{array}$$

(12)

Proof. 

By Lemma 1 we have

$$\sum _{k=0}^{\infty }\frac{1}{x^k\left(\genfrac{}{}{0pt}{}{3k}{k}\right)}={\int }_0^1\frac{-x(2t^3-2t^2-x)}{{(t^3-t^2+x)}^2}dt.$$

By equation (3) we have $\left|x\right|>\frac{4}{27}$. □

Let $x=-2b{(1+2b)}^2$, then we have $t^3-t^2+x=(t-(2b+1))({(t+b)}^2+b(3b+2))$. If $\left|x\right|>\frac{4}{27}$, we have $b<-\frac{2}{3}$ or $b>\mu$. If $b\ne -\frac{3}{4}$, we have

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }\frac{1}{x^k\left(\genfrac{}{}{0pt}{}{3k}{k}\right)}\hfill \\ \hfill =& {\int }_0^1\frac{-x(2t^3-2t^2-x)}{{(t^3-t^2+x)}^2}dt\hfill \\ \hfill =& {\int }_0^1\frac{2b{(1+2b)}^2(2t^3-2t^2+2b{(1+2b)}^2)}{{(t-(2b+1))}^2{({(t+b)}^2+b(3b+2))}^2}dt\hfill \end{array}$$

We will calculate the integral by Mathematica. Then we get

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }\frac{1}{{(-2b{(1+2b)}^2)}^k\left(\genfrac{}{}{0pt}{}{3k}{k}\right)}\hfill \\ \hfill =& \frac{\sqrt{b}}{{(2+3b)}^{3/2}}\frac{1+2b}{{(1+6b)}^3}[27\sqrt{b}\sqrt{2+3b}(2b+1)(6b+1)\hfill \\ \hfill +& 6\sqrt{b}\sqrt{2+3b}(2+3b)log\left(\frac{2b}{1+b}\right)\hfill \\ \hfill +& 2(18b^2+6b-1)\left(arctan\frac{1+b}{\sqrt{b}\sqrt{2+3b}}-arctan\left(\frac{\sqrt{b}}{\sqrt{2+3b}}\right)\right)]\hfill \end{array}$$

Let

$$q\left(b\right)=arctan\frac{1+b}{\sqrt{b}\sqrt{2+3b}}-arctan\left(\frac{\sqrt{b}}{\sqrt{2+3b}}\right),$$

It’s easy to deduce that

$$q\left(b\right)=\left\{\begin{array}{cc}arctan\left(\sqrt{\frac{2+3b}{b}}\frac{1}{4b+3}\right),\hfill & b<-\frac{3}{4},orb>0\hfill \\ arctan\left(\sqrt{\frac{2+3b}{b}}\frac{1}{4b+3}\right)-\pi ,\hfill & -\frac{3}{4}<b<-\frac{2}{3}\hfill \end{array}\right.$$

In the domain of b we have

$$\frac{\sqrt{b}}{{(2+3b)}^{3/2}}\sqrt{b}\sqrt{2+3b}=\frac{b}{2+3b}$$

and

$$\frac{\sqrt{b}}{\sqrt{2+3b}}=\sqrt{\frac{b}{2+3b}}.$$

If $b=-\frac{3}{4}$, the theorem also holds, so we complete the proof.

Using the same method of Theorem 3 and by Lemma 2 we have

Theorem 4. 

Let μ be defined by

$$\mu =\frac{1}{6}\left({(1+\sqrt{2})}^{2/3}+\frac{1}{{(1+\sqrt{2})}^{2/3}}\right)-\frac{1}{3}.$$

If we define $q\left(b\right)$ as

$$q\left(b\right)=\left\{\begin{array}{cc}arctan\left(\sqrt{\frac{2+3b}{b}}\frac{1}{4b+3}\right),\hfill & b<-\frac{3}{4},orb>\mu \hfill \\ -\frac{\pi }{2},\hfill & b=-\frac{3}{4}\hfill \\ arctan\left(\sqrt{\frac{2+3b}{b}}\frac{1}{4b+3}\right)-\pi ,\hfill & -\frac{3}{4}<b<-\frac{3}{2}\hfill \end{array}\right.$$

If $b>\mu$ or $b<-\frac{2}{3}$, we have

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }\frac{1}{(1+3k)\left(\genfrac{}{}{0pt}{}{3k}{k}\right){(-2b{(1+2b)}^2)}^k}\hfill \\ \hfill =& -\frac{3b(1+2b)}{1+6b}log\left(\frac{2b}{1+2b}\right)+\frac{2(1+2b)(1+3b)}{(1+6b)}\sqrt{\frac{b}{2+3b}}q\left(b\right).\hfill \end{array}$$

(13)

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }\frac{k}{{(-2b{(1+2b)}^2)}^k\left(\genfrac{}{}{0pt}{}{3k}{k}\right)}\hfill \\ \hfill =& -\frac{81b{(1+2b)}^2}{{(2+3b)}^2{(1+6b)}^4}+\frac{6b(1+2b)(-1+8b+12b^2)}{{(1+6b)}^5}log\left(\frac{2b}{1+2b}\right)\hfill \\ \hfill +& \frac{2(1+2b)f\left(b\right)}{{(2+3b)}^2{(1+6b)}^5}\sqrt{\frac{b}{2+3b}}q\left(b\right).\hfill \end{array}$$

(14)

where $f\left(b\right)=1-45b-150b^2+54b^3+648b^4+648b^5$.

$$\begin{array}{cc}\hfill & \sum _{k=1}^{\infty }\frac{1}{k{(-2b{(1+2b)}^2)}^k\left(\genfrac{}{}{0pt}{}{3k}{k}\right)}\hfill \\ \hfill =& \frac{1}{1+6b}log\left(\frac{2b}{1+2b}\right)+\frac{2}{1+6b}\sqrt{\frac{b}{2+3b}}q\left(b\right).\hfill \end{array}$$

(15)

Remark 1. 

If $b<0$, $\sqrt{b}$ is defined by $\sqrt{b}=\sqrt{\left|b\right|}i$

By Theorem 3 and 4 we can get propositions of equality.

Proposition 3. 

Let μ be defined by

$$\mu =\frac{1}{6}\left({(1+\sqrt{2})}^{2/3}+\frac{1}{{(1+\sqrt{2})}^{2/3}}\right)-\frac{1}{3}.$$

If we define $q\left(b\right)$ as

$$q\left(b\right)=\left\{\begin{array}{cc}arctan\left(\sqrt{\frac{2+3b}{b}}\frac{1}{4b+3}\right),\hfill & b<-\frac{3}{4},orb>\mu \hfill \\ -\frac{\pi }{2},\hfill & b=-\frac{3}{4}\hfill \\ arctan\left(\sqrt{\frac{2+3b}{b}}\frac{1}{4b+3}\right)-\pi ,\hfill & -\frac{3}{4}<b<-\frac{3}{2}\hfill \end{array}\right.$$

If $b>\mu$ or $b<-\frac{2}{3}$, we have:

Via $\left(13\right)+3b(1+2b)\times \left(15\right)$ we get

$$\sum _{k=1}^{\infty }\frac{(18b^2+9b+1)k+3b(1+2b)}{k(1+3k){(-2b{(1+2b)}^2)}^k\left(\genfrac{}{}{0pt}{}{3k}{k}\right)}=2(1+2b)\sqrt{\frac{b}{3+2b}}q\left(b\right)-1$$

(16)

Via ${(1+6b)}^2\times \left(12\right)+2\times \left(13\right)$ we get

$$\sum _{k=0}^{\infty }\frac{{(1+6b)}^{2}k+(1+4b+12b^2)}{(1+3k){(-2b{(1+2b)}^2)}^k\left(\genfrac{}{}{0pt}{}{3k}{k}\right)}=\frac{9b{(1+2b)}^2}{2+3b}+\frac{2{(1+2b)}^2}{2+3b}\sqrt{\frac{b}{2+3b}}q\left(b\right)$$

(17)

By Proposition 3, by equality (16) and (17) we let $b=-1$ and $b=-\frac{3}{4}$, we can get four $\pi$-series, here are the examples.

Example 3. 

$$\sum _{k=1}^{\infty }\frac{3+10k}{k(1+3k)2^k\left(\genfrac{}{}{0pt}{}{3k}{k}\right)}=-1+\frac{\pi }{2}$$

(18)

$$\sum _{k=1}^{\infty }\frac{(9+35k)8^{k-1}}{k(1+3k)3^k\left(\genfrac{}{}{0pt}{}{3k}{k}\right)}=\frac{\sqrt{3}}{2}\pi -1$$

(19)

$$\sum _{k=0}^{\infty }\frac{9+25k}{(1+3k)2^k\left(\genfrac{}{}{0pt}{}{3k}{k}\right)}=\frac{\pi }{2}+9$$

(20)

$$\sum _{k=0}^{\infty }\frac{(19+49k)8^k}{4(1+3k)3^k\left(\genfrac{}{}{0pt}{}{3k}{k}\right)}=\frac{27}{4}+\sqrt{3}\pi$$

(21)

4. $\mathbf{\pi }$-series containing $\mathbf{1}/\left(\genfrac{}{}{0pt}{}{\mathbf{4}\mathbf{k}}{\mathbf{k}}\right)$

In this section, we will give methods to find $\pi$-series containing the type $1/\left(\genfrac{}{}{0pt}{}{4k}{k}\right)$.

We first give an auxiliary lemma we will use.

Lemma 4. 

Let $f\left(t\right)=t^4-t^3+x$, let $x=\frac{8b^3{(1+2b)}^3}{{(1+4b)}^2}$, if $x>\frac{27}{256}$, then we can factor $f\left(t\right)$ as

$$f\left(t\right)=\left({\left(t-b-\frac{1}{2}\right)}^2+\frac{{(1+2b)}^2(-1+4b)}{4(1+4b)}\right)\left({(t+b)}^2+\frac{b^2(3+4b)}{1+4b}\right),$$

where $b>\frac{1}{4}$ or $b<-\frac{3}{4}$.

If $x<-\frac{27}{256}$, then we can factor $f\left(t\right)$ as

$$f\left(t\right)=\left(t^2-(1+2b)t+\frac{2b{(1+2b)}^2}{1+4b}\right)\left({(t+b)}^2+\frac{b^2(3+4b)}{1+4b}\right),$$

where $-0.33\approx -\frac{1}{4}-\mu <b<-\frac{1}{4}+\mu \approx -0.17$ and $b\ne -\frac{1}{4}$ and

$$\mu =\frac{\sqrt{2}}{8}\sqrt{\frac{-3{(1+\sqrt{2})}^{2/3}+2{(1+\sqrt{2})}^{1/3}+3}{{(1+\sqrt{2})}^{1/3}}}\approx 0.08.$$

Proof. 

Consider $f\left(t\right)=0$, since $\left|x\right|>\frac{27}{256}$, we can deduce that

$$\Delta =-12288x^2(256x-27)\left\{\begin{array}{c}<0,x>\frac{27}{256}\hfill \\ >0,x<-\frac{27}{256}\hfill \end{array}\right.$$

If $x>\frac{27}{256}$, then $f\left(t\right)=0$ has two pairs of unequal conjugate complex roots. Suppose we factor $f\left(t\right)$ as

$$f\left(t\right)=({(t+a)}^2+c)({(t+b)}^2+d),$$

Then we have

$$\left\{\begin{array}{c}2(a+b)=-1\hfill \\ a^2+b^2+4ab+c+d=0\hfill \\ 2(a^2+c)b+2(b^2+d)a=0\hfill \\ (a^2+c)(b^2+d)=x\hfill \end{array}\right.$$

That is to say $a=-\frac{1}{2}-b$ and $c=\frac{{(1+2b)}^2(-1+4b)}{4(1+4b)}$ and $d=\frac{b^2(3+4b)}{1+4b}$ and $x=\frac{8b^3{(1+2b)}^3}{{(1+4b)}^2}$. If $x>\frac{4}{27}$, we have $b<-\frac{3}{4}$ or $b>\frac{1}{4}$. □

If $x<-\frac{27}{256}$, then $f\left(t\right)=0$ has two unequal real roots and a pair of conjugate complex roots. Suppose we factor $f\left(t\right)$ as

$$f\left(t\right)=(t+a)(t+c)({(t+b)}^2+d)=(t+(a+c)t+ac)({(t+b)}^2+d),$$

Then we have

$$\left\{\begin{array}{c}a+c+2b=-1\hfill \\ ac+2(a+c)b+b^2+d=0\hfill \\ 2acb+(a+c)(b^2+d)=0\hfill \\ ac(b^2+d)=x\hfill \\ a\ne c\hfill \end{array}\right.$$

That is to say $a+c=-(1+2b)$ and $ac=\frac{2b{(1+2b)}^2}{1+4b}$ and $d=\frac{b^2(3+4b)}{1+4b}$ and $x=\frac{8b^3{(1+2b)}^3}{{(1+4b)}^2}$. Since $x<-\frac{27}{256}$ and $a\ne c$, we have $-\frac{1}{4}-\mu <b<-\frac{1}{4}+\mu$ and $b\ne -\frac{1}{4}$.

Theorem 5. 

Let

$$\mu =\frac{\sqrt{2}}{8}\sqrt{\frac{-3{(1+\sqrt{2})}^{2/3}+2{(1+\sqrt{2})}^{1/3}+3}{{(1+\sqrt{2})}^{1/3}}}.$$

If $b>\frac{1}{4}$ or $b<-\frac{3}{4}$ or $-\frac{1}{4}-\mu <b<-\frac{1}{4}+\mu$ and $b\ne -\frac{1}{4}$, then we have

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }\frac{{(1+4b)}^{2k}}{{\left(8b^3{(1+2b)}^3\right)}^k\left(\genfrac{}{}{0pt}{}{4k}{k}\right)}\hfill \\ \hfill =& \frac{2048b^3{(1+2b)}^3}{(-1+4b)(3+4b){(3+16b+32b^2)}^2}\hfill \\ \hfill +& \frac{6b(1+2b)\sqrt{(-1+4b)(1+4b)}{f}_1\left(b\right)}{{(-1+4b)}^2{(3+16b+32b^2)}^3}arctan\left(\frac{\sqrt{(-1+4b)(1+4b)}}{-1+8b^2}\right)\hfill \\ \hfill +& \frac{6b(1+2b)\sqrt{(1+4b)(3+4b)}{f}_2\left(b\right)}{{(3+4b)}^2{(3+16b+32b^2)}^3}arctan\left(\frac{\sqrt{(1+4b)(3+4b)}}{1+8b+8b^2}\right)\hfill \\ \hfill +& \frac{6b(1+2b)(1+4b)(1+16b+32b^2)}{{(3+16b+32b^2)}^3}log\left(\frac{4b^2}{{(1+2b)}^2}\right)\hfill \end{array}$$

(22)

where

$$f_1\left(b\right)=1+12b+176b^2+640b^3+512b^4$$

and

$$f_2\left(b\right)=-9-60b-16b^2+384b^3+512b^4.$$

Proof. 

By Lemma 1 we have

$$\sum _{k=0}^{\infty }\frac{1}{x^k\left(\genfrac{}{}{0pt}{}{4k}{k}\right)}={\int }_0^1\frac{-x(3t^4-3t^3-x)}{{(t^4-t^3+x)}^2}dt.$$

By equation (3) we have $\left|x\right|>\frac{27}{256}$. □

Let $x=\frac{8b^3{(1+2b)}^3}{{(1+4b)}^2}$. Since $\left|x\right|>\frac{27}{256}$, we have $b>\frac{1}{4}$ or $b<-\frac{3}{4}$ or $-\frac{1}{4}-\mu <b<-\frac{1}{4}+\mu$ and $b\ne -\frac{1}{4}$. By Mathematica we have

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }\frac{1}{x^k\left(\genfrac{}{}{0pt}{}{4k}{k}\right)}\hfill \\ \hfill =& {\int }_0^1\frac{-x(3t^4-3t^3-x)}{{(t^4-t^3+x)}^2}dt\hfill \\ \hfill =& \frac{b^3{(1+2b)}^3}{{(1+4b)}^2{(3+16b+32b^2)}^3}[\frac{2048{(1+4b)}^2(3+16b+32b^2)}{-3+8b+16b^2}\hfill \\ \hfill +& \frac{6{(1+4b)}^2{f}_1\left(b\right)}{b^2{(1+2b)}^2(-1+4b)}\frac{\sqrt{1+4b}}{\sqrt{-1+4b}}\left(arctan\frac{1}{\sqrt{\frac{-1+4b}{1+4b}}}-arctan\frac{-1+2b}{(1+2b)\sqrt{\frac{-1+4b}{1+4b}}}\right)\hfill \\ \hfill +& \frac{6{(1+4b)}^2{f}_2\left(b\right)}{b^2{(1+2b)}^2(3+4b)}\frac{\sqrt{1+4b}}{\sqrt{3+4b}}\left(arctan\frac{(1+b)\sqrt{\frac{1+4b}{3+4b}}}{b}-arctan\sqrt{\frac{1+4b}{3+4b}}\right)\hfill \\ \hfill +& \frac{6{(1+4b)}^3(1+16b+32b^2)}{b^2{(1+2b)}^2}log\left(\frac{4b^2}{{(1+2b)}^2}\right)].\hfill \end{array}$$

where

$$f_1\left(b\right)=1+12b+176b^2+640b^3+512b^4$$

and

$$f_2\left(b\right)=-9-60b-16b^2+384b^3+512b^4.$$

Since in the domain of b we have

$$\begin{array}{cc}\hfill & \frac{\sqrt{1+4b}}{\sqrt{-1+4b}}\left(arctan\frac{1}{\sqrt{\frac{-1+4b}{1+4b}}}-arctan\frac{-1+2b}{(1+2b)\sqrt{\frac{-1+4b}{1+4b}}}\right)\hfill \\ \hfill =& \frac{\sqrt{(-1+4b)(1+4b)}}{-1+4b}arctan\left(\frac{\sqrt{(-1+4b)(1+4b)}}{-1+8b^2}\right)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & \frac{\sqrt{1+4b}}{\sqrt{3+4b}}\left(arctan\frac{(1+b)\sqrt{\frac{1+4b}{3+4b}}}{b}-arctan\sqrt{\frac{1+4b}{3+4b}}\right)\hfill \\ \hfill =& \frac{\sqrt{(1+4b)(3+4b)}}{3+4b}arctan\left(\frac{\sqrt{(1+4b)(3+4b)}}{1+8b+8b^2}\right).\hfill \end{array}$$

Then we get the desired results.

Using the same method of Theorem 5 and by Lemma 2 we have

Theorem 6. 

Let

$$\mu =\frac{\sqrt{2}}{8}\sqrt{\frac{-3{(1+\sqrt{2})}^{2/3}+2{(1+\sqrt{2})}^{1/3}+3}{{(1+\sqrt{2})}^{1/3}}}.$$

If $b>\frac{1}{4}$ or $b<-\frac{3}{4}$ or $-\frac{1}{4}-\mu <b<-\frac{1}{4}+\mu$ and $b\ne -\frac{1}{4}$, then we have

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }\frac{{(1+4b)}^{2k}}{{\left(8b^3{(1+2b)}^3\right)}^k(4k+1)\left(\genfrac{}{}{0pt}{}{4k}{k}\right)}\hfill \\ \hfill =& \frac{2b(1+2b)(1+4b+16b^2)\sqrt{(-1+4b)(1+4b)}}{(-1+4b)(1+4b)(3+16b+32b^2)}arctan\left(\frac{\sqrt{(-1+4b)(1+4b)}}{8b^2-1}\right)\hfill \\ \hfill +& \frac{2b(1+2b)(3+12b+16b^2)\sqrt{(1+4b)(3+4b)}}{(1+4b)(3+4b)(3+16b+32b^2)}arctan\left(\frac{\sqrt{(1+4b)(3+4b)}}{1+8b+8b^2}\right)\hfill \\ \hfill -& \frac{2b(1+2b)(1+4b)}{3+16b+32b^2}log\left(\frac{4b^2}{{(1+2b)}^2}\right).\hfill \end{array}$$

(23)

$$\begin{array}{cc}\hfill & \sum _{k=1}^{\infty }\frac{{(1+4b)}^{2k}}{{\left(8b^3{(1+2b)}^3\right)}^{k}k\left(\genfrac{}{}{0pt}{}{4k}{k}\right)}\hfill \\ \hfill =& \frac{3(1+2b)\sqrt{(-1+4b)(1+4b)}}{(-1+4b)(3+16b+32b^2)}arctan\left(\frac{\sqrt{(-1+4b)(1+4b)}}{-1+8b^2}\right)\hfill \\ \hfill +& \frac{6b\sqrt{(1+4b)(3+4b)}}{(3+4b)(3+16b+32b^2)}arctan\left(\frac{\sqrt{(1+4b)(3+4b)}}{1+8b^{+}8b^2}\right)\hfill \\ \hfill +& \frac{3(1+4b)}{2(3+16b+32b^2)}log\left(\frac{4b^2}{{(1+2b)}^2}\right).\hfill \end{array}$$

(24)

By Theorem 5 and 6 we can get propositions of equality.

Proposition 4. 

Let

$$\mu =\frac{\sqrt{2}}{8}\sqrt{\frac{-3{(1+\sqrt{2})}^{2/3}+2{(1+\sqrt{2})}^{1/3}+3}{{(1+\sqrt{2})}^{1/3}}}.$$

If $b>\frac{1}{4}$ or $b<-\frac{3}{4}$ or $-\frac{1}{4}-\mu <b<-\frac{1}{4}+\mu$ and $b\ne -\frac{1}{4}$, then we have:

Via $(3+4b){(3+16b+32b^2)}^2\times \left(22\right)+3(1+4b)(3+8b)\times \left(23\right)-128b^2{(1+2b)}^3\times \left(24\right)$ we have

$$\begin{array}{cc}\hfill & \frac{[(3+4b){(3+16b+32b^2)}^2{k}^2+f_1\left(b\right)k-32b^2{(1+2b)}^3]{(1+4b)}^{2k}}{k(1+4k){\left(8b^3{(1+2b)}^3\right)}^k\left(\genfrac{}{}{0pt}{}{4k}{k}\right)}\hfill \\ \hfill =& \frac{3\left(16b^2(1+2b^2)\sqrt{16b^2-1}arctan\frac{\sqrt{16b^2-1}}{-1+8b^2}-f_2\left(b\right)\right)}{{(4b-1)}^2}\hfill \end{array}$$

(25)

where

$$f_1\left(b\right)=9+96b+328b^2+448b^3+256b^4$$

and

$$f_2\left(b\right)=(-1+4b)(1+4b)(-3-8b+8b^2).$$

Via $(-1+4b){(3+16b+32b^2)}^2\times \left(22\right)-3(1+4b)(1+8b)\times \left(23\right)-256b^3{(1+2b)}^2\times \left(24\right)$ we have

$$\begin{array}{cc}\hfill & \frac{[(-1+4b){(3+16b+32b^2)}^2{k}^2-f_3\left(b\right)k-64b^3{(1+2b)}^2]{(1+4b)}^{2k}}{k(1+4k){\left(8b^3{(1+2b)}^3\right)}^k\left(\genfrac{}{}{0pt}{}{4k}{k}\right)}\hfill \\ \hfill =& \frac{3\left(f_4\left(b\right)-16b^2{(1+2b)}^2\sqrt{(1+4b)(3+4b)}arctan\frac{\sqrt{(1+4b)(3+4b)}}{1+8b+8b^2}\right)}{{(3+4b)}^2}\hfill \end{array}$$

(26)

where

$$f_3\left(b\right)=3+24b+40b^2+64b^3+256b^4$$

and

$$f_2\left(b\right)=(1+4b)(3+4b)(3+16b+8b^2).$$

By Proposition 4, in equation (25) we let $b=\frac{1}{2}$ and in (26) we let $b=-1$, we can get one $\pi$-series, here is the example.

Example 4. 

$$\sum _{k=1}^{\infty }\frac{(-64+211k+1805k^2)9^k}{k(1+4k)8^k\left(\genfrac{}{}{0pt}{}{4k}{k}\right)}=45+16\sqrt{3}\pi$$

(27)

5. $\mathbf{\pi }$-series containing $\mathbf{1}/\left(\genfrac{}{}{0pt}{}{\mathbf{4}\mathbf{k}}{\mathbf{2}\mathbf{k}}\right)$

In this section, we will give methods to find $\pi$-series containing the type $1/\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)$.

We first give an auxiliary lemma we will use.

Lemma 5. 

Let $f\left(t\right)=t^4-2t^3+t^2-x$, if $x<-\frac{1}{16}$, let $x=-b(1+b){(1+2b)}^2$, then we can factor $f\left(t\right)$ as

$$f\left(t\right)=({\left(t-b-1\right)}^2+b(1+b))\left({(t+b)}^2+b(1+b)\right),$$

where $b>\frac{\sqrt{2+2\sqrt{2}}}{4}-\frac{1}{2}\approx 0.0493$ or $b<-\frac{1}{2}-\frac{2+2\sqrt{2}}{4}\approx -1.05$.

If $x>\frac{1}{16}$, let $x=b^2$, then we can factor $f\left(t\right)$ as

$$f\left(t\right)=\left({(t-1/2)}^2+b-1/4\right)\left({(t-1/2)}^2-b-1/4\right),$$

where $\left|b\right|>1/4$.

Proof. 

Consider $f\left(t\right)=0$, since $\left|x\right|>\frac{1}{16}$, we can deduce that

$$\Delta =196608x^2(16x-1)\left\{\begin{array}{c}>0,x>\frac{1}{16}\hfill \\ <0,x<-\frac{1}{16}\hfill \end{array}\right.$$

If $x>\frac{1}{16}$, then $f\left(t\right)=0$ has two unequal real roots and a pair of conjugate complex roots. We then get

$$\begin{array}{cc}\hfill f\left(t\right)=& t^4-2t^3+t^2-b^2\hfill \\ \hfill =& t^2{(t-1)}^2-b^2\hfill \\ \hfill =& \left(t\right(t-1)-b)\left(t\right(t-1)+b)\hfill \\ \hfill =& ({(t-1/2)}^2-b-1/4)({(t-1/2)}^2+b-1/4)\hfill \end{array}$$

□

If $x<-\frac{1}{16}$, then $f\left(t\right)$ has two pair of conjugate complex roots. Suppose we factor $f\left(t\right)$ as

$$f\left(t\right)=({(t+a)}^2+c)({(t+b)}^2+d),$$

Then we have

$$\left\{\begin{array}{c}2(a+b)=-2\hfill \\ a^2+b^2+4ab+c+d=1\hfill \\ 2(a^2+c)b+2(b^2+d)a=0\hfill \\ (a^2+c)(b^2+d)=x\hfill \end{array}\right.$$

That is to say $a=-(b+1)$ and $c=b(1+b)$ and $d=b(1+b)$ and $x=-b(1+b){(1+2b)}^2$. Since $x<-\frac{1}{16}$, we have $b>\frac{\sqrt{2+2\sqrt{2}}}{4}-\frac{1}{2}\approx 0.0493$ or $b<-\frac{1}{2}-\frac{2+2\sqrt{2}}{4}\approx -1.05$

We first consider the case $x<-\frac{1}{16}$, then we have the following theorem.

Theorem 7. 

If $b>\frac{\sqrt{2+2\sqrt{2}}}{4}-\frac{1}{2}\approx 0.0493$ or $b<-\frac{1}{2}-\frac{2+2\sqrt{2}}{4}\approx -1.05$, then we have

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }\frac{1}{{(-b(1+b){(1+2b)}^2)}^k\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}\hfill \\ \hfill =& \frac{16b(1+b)(1+2b)(1+10b+24b^2+16b^3)}{{(1+8b+8b^2)}^3}\hfill \\ \hfill +& \frac{2{(1+2b)}^2(-1+8b+8b^2)\sqrt{b(1+b)}}{{(1+8b+8b^2)}^3}arctan\frac{1}{2\sqrt{b(1+b)}}\hfill \\ \hfill +& \frac{2b(1+b)(1+2b)(3+8b+8b^2)}{{(1+8b+8b^2)}^3}log\left(\frac{b}{1+b}\right)\hfill \end{array}$$

(28)

Proof. 

By Lemma 1 we have

$$\sum _{k=0}^{\infty }\frac{1}{x^k\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}={\int }_0^1\frac{x(3t^2{(1-t)}^2+x)}{{(t^2{(1-t)}^2-x)}^2}dt.$$

By inequality (3) we have $\left|x\right|>\frac{1}{16}$.

Let $x=-b(1+b){(1+2b)}^2$, then we have

$$f\left(t\right)=({\left(t-b-1\right)}^2+b(1+b))\left({(t+b)}^2+b(1+b)\right),$$

If $x<-\frac{1}{16}$, we have $b>\frac{\sqrt{2+2\sqrt{2}}}{4}-\frac{1}{2}\approx 0.0493$ or $b<-\frac{1}{2}-\frac{2+2\sqrt{2}}{4}\approx -1.05$, then by Mathematica we get

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }\frac{1}{x^k\left(\genfrac{}{}{0pt}{}{4k}{k}\right)}\hfill \\ \hfill =& {\int }_0^1\frac{x(3t^2{(1-t)}^2+x)}{{(t^2{(1-t)}^2-x)}^2}dt\hfill \\ \hfill =& \frac{2(1+2b)\sqrt{b}\sqrt{1+b}}{{(1+8b+8b^2)}^3}[8\sqrt{b}\sqrt{1+b}(1+10b+24b^2+16b^3)\hfill \\ \hfill +& (1+2b)(-1+8b+8b^2)\left(arctan\sqrt{\frac{1+b}{b}}-arctan\sqrt{\frac{b}{1+b}}\right)\hfill \\ \hfill +& (3+8b+8b^2)\sqrt{b}\sqrt{1+b}log\left(\frac{b}{1+b}\right)]\hfill \end{array}$$

Since in the domain of b we have

$$\sqrt{b}\sqrt{1+b}\sqrt{b}\sqrt{1+b}=b(1+b)$$

and

$$\begin{array}{cc}\hfill & \sqrt{b}\sqrt{1+b}\left(arctan\sqrt{\frac{1+b}{b}}-arctan\sqrt{\frac{b}{1+b}}\right)\hfill \\ \hfill =& \sqrt{b(1+b)}arctan\frac{1}{2\sqrt{b(1+b)}}.\hfill \end{array}$$

Then by some simplification we get the desired results. □

Using the same method of Theorem 7 and by Lemma 2 we have

Theorem 8. 

If $b>\frac{\sqrt{2+2\sqrt{2}}}{4}-\frac{1}{2}\approx 0.0493$ or $b<-\frac{1}{2}-\frac{2+2\sqrt{2}}{4}\approx -1.05$, then we have

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }\frac{1}{{(-b(1+b){(1+2b)}^2)}^k(4k+1)\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}\hfill \\ \hfill =& \frac{2{(1+2b)}^2\sqrt{b(1+b)}}{1+8b+8b^2}arctan\frac{1}{2\sqrt{b(1+b)}}\hfill \\ \hfill -& \frac{2b(1+b)(1+2b)}{1+8b+8b^2}log\left(\frac{b}{1+b}\right)\hfill \end{array}$$

(29)

$$\begin{array}{cc}\hfill & \sum _{k=1}^{\infty }\frac{1}{{(-b(1+b){(1+2b)}^2)}^{k}k\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}\hfill \\ \hfill =& \frac{4\sqrt{b(1+b)}}{1+8b+8b^2}arctan\frac{1}{2\sqrt{b(1+b)}}\hfill \\ \hfill +& \frac{1+2b}{1+8b+8b^2}log\left(\frac{b}{1+b}\right)\hfill \end{array}$$

(30)

By Theorem 7 and 8 we get some propositions of equality.

Proposition 5. 

If $b>\frac{\sqrt{2+2\sqrt{2}}}{4}-\frac{1}{2}$ or $b<-\frac{1}{2}-\frac{2+2\sqrt{2}}{4}$, then we have:

Via $\left(29\right)+2b(1+2b)\times \left(30\right)$ we get

$$\sum _{k=1}^{\infty }\frac{(1+8b+8b^2)k+2b(1+b)}{k(4k+1){(-b(1+b){(1+2b)}^2)}^k\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}=2\sqrt{b(1+b)}arctan\frac{1}{2\sqrt{b(1+b)}}-1$$

(31)

Via ${(1+8b+8b^2)}^2\times \left(28\right)-2b(1+b)(3+8b+8b^2)\times \left(30\right)$ we get

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }\frac{{(1+8b+8b^2)}^{2}k-2b(1+b)(3+8b+8b^2)}{k{(-b(1+b){(1+2b)}^2)}^k\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}\hfill \\ \hfill =& -\left(1+2\sqrt{b(1+b)}arctan\frac{1}{2\sqrt{b(1+b)}}\right)\hfill \end{array}$$

(32)

By Proposition 5 and in equation (31) and (32) we let $b=\frac{1}{2},\frac{\sqrt{2}-1}{2},\frac{\sqrt{3}}{3}-\frac{1}{2}$, we can get some $\pi$-series, here are the examples.

Example 5. 

$$\sum _{k=1}^{\infty }\frac{3+14k}{2k(1+4k){(-3)}^k\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}=\frac{1}{6}(-6+\sqrt{3}\pi )$$

(33)

$$\sum _{k=1}^{\infty }\frac{(1+6k){(-2)}^k}{2k(1+4k)\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}=\frac{1}{4}(-4+\pi )$$

(34)

$$\sum _{k=1}^{\infty }\frac{(1+10k){(-9)}^k}{6k(1+4k)\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}=\frac{1}{9}(-9+\sqrt{3}\pi )$$

(35)

$$\sum _{k=1}^{\infty }\frac{-27+98k}{2k{(-3)}^k\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}=-\left(1+\frac{\pi }{2\sqrt{3}}\right)$$

(36)

$$\sum _{k=1}^{\infty }\frac{(-5+18){(-2)}^k}{2k\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}=-\left(1+\frac{\pi }{4}\right)$$

(37)

$$\sum _{k=1}^{\infty }\frac{(-11+50k){(-9)}^k}{18k\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}=-\left(1+\frac{\pi }{3\sqrt{3}}\right)$$

(38)

Now we consider the case $x>\frac{1}{16}$.

Theorem 9. 

If $\left|b\right|>\frac{1}{4}$, then we have

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }\frac{1}{b^{2k}\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}\hfill \\ \hfill =& \frac{16b^2}{16b^2-1}+\frac{2b}{\sqrt{-1-4b}(1+4b)}arctan\frac{1}{\sqrt{-1-4b}}\hfill \\ \hfill +& \frac{2b}{(-1+4b)\sqrt{-1+4b}}arctan\frac{1}{\sqrt{-1+4b}}.\hfill \end{array}$$

(39)

Proof. 

By Lemma 1 we have

$$\sum _{k=0}^{\infty }\frac{1}{x^k\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}={\int }_0^1\frac{x(3t^2{(1-t)}^2+x)}{{(t^2{(1-t)}^2-x)}^2}dt.$$

By equation (3) we have $\left|x\right|>\frac{1}{16}$. □

Let $x=b^2$, then we have

$$f\left(t\right)=({(t-1/2)}^2-b-1/4)({(t-1/2)}^2+b-1/4).$$

If $x>\frac{1}{16}$, we have $\left|b\right|>\frac{1}{4}$ then by Mathematica we have

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }\frac{1}{b^{2k}\left(\genfrac{}{}{0pt}{}{4k}{k}\right)}\hfill \\ \hfill =& {\int }_0^1\frac{b^2(3t^2{(1-t)}^2+b^2)}{{({(t-1/2)}^2-b-1/4)}^2{({(t-1/2)}^2+b-1/4)}^2}dt\hfill \\ \hfill =& \frac{-2b}{{(-1-4b)}^{3/2}{(-1+4b)}^{3/2}}[8b\sqrt{-1-4b}\sqrt{-1+4b}\hfill \\ \hfill +& {(-1+4b)}^{3/2}arctan\frac{1}{\sqrt{-1-4b}}-{(-1-4b)}^{3/2}arctan\frac{1}{\sqrt{-1+4b}}].\hfill \end{array}$$

Then by some simplifications we get the desired results.

Using the same method of Theorem 9 and by Lemma 2 we have

Theorem 10. 

If $\left|b\right|>\frac{1}{4}$, then we have

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }\frac{1}{(4k+1)b^{2k}\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}\hfill \\ \hfill =& 2b\left(\frac{arctan\frac{1}{\sqrt{-1+4b}}}{\sqrt{-1+4b}}-\frac{arctan\frac{1}{\sqrt{-1-4b}}}{\sqrt{-1-4b}}\right).\hfill \end{array}$$

(40)

$$\begin{array}{cc}\hfill & \sum _{k=0}^{\infty }\frac{k}{b^{2k}\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}=\frac{24b^2}{{(-1+4b)}^2{(1+4b)}^2}\hfill \\ \hfill +& \frac{b(-1+2b)}{{(1+4b)}^2}\frac{arctan\frac{1}{\sqrt{-1-4b}}}{\sqrt{-1-4b}}+\frac{b(1+2b)}{{(-1+4b)}^2}\frac{arctan\frac{1}{\sqrt{-1+4b}}}{\sqrt{-1+4b}}.\hfill \end{array}$$

(41)

$$\begin{array}{c}\hfill \sum _{k=1}^{\infty }\frac{1}{k{b}^{2k}\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}=2\left(\frac{arctan\frac{1}{\sqrt{-1-4b}}}{\sqrt{-1-4b}}+\frac{arctan\frac{1}{\sqrt{-1+4b}}}{\sqrt{-1+4b}}\right).\end{array}$$

(42)

By Theorem 10 we can get propositions of equality.

Proposition 6. 

If $\left|b\right|>\frac{1}{4}$, via $\left(40\right)+b\times \left(42\right)$ we have

$$\sum _{k=1}^{\infty }\frac{(4b+1)k+b}{k(4k+1)b^{2k}\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}=\frac{4b}{\sqrt{4b-1}}arctan\frac{1}{\sqrt{4b-1}}-1.$$

(43)

If $\left|b\right|>\frac{1}{4}$, via $(-1+2b)\times \left(40\right)+2{(1+4b)}^2\times \left(41\right)$ we have

$$\sum _{k=0}^{\infty }\frac{2{(1+4b)}^{2}k(4k+1)+2b-1}{(4k+1)b^{2k}\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}=\frac{48b^2}{{(-1+4b)}^2}+\frac{8b^2(5+16b^2)}{{(-1+4b)}^2}\frac{arctan\frac{1}{\sqrt{-1+4b}}}{\sqrt{-1+4b}}.$$

(44)

By Proposition 6 and in equation (43) and (44) we let $b=\frac{1}{3},\frac{1}{2},1$, we can get some $\pi$-series, here are the examples.

Example 6. 

$$\sum _{k=1}^{\infty }\frac{(1+7k)3^{2k-1}}{k(1+4k)\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}=\frac{4\sqrt{3}\pi }{9}-1$$

(45)

$$\sum _{k=1}^{\infty }\frac{1+5k}{k(1+4k)\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}=\frac{2\sqrt{3}\pi }{9}-1$$

(46)

$$\sum _{k=1}^{\infty }\frac{(1+6k)2^{2k-1}}{k(1+4k)\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}=\frac{\pi }{2}-1$$

(47)

$$\sum _{k=1}^{\infty }\frac{(-3+98k+392k^2)9^{k-1}}{(1+4k)\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}=\frac{488\pi }{9\sqrt{3}}+48$$

(48)

$$\sum _{k=1}^{\infty }\frac{1+50k+200k^2}{(1+4k)\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}=\frac{4}{27}(36+7\sqrt{3}\pi )$$

(49)

$$\sum _{k=1}^{\infty }\frac{k{4}^k}{\left(\genfrac{}{}{0pt}{}{4k}{2k}\right)}=\frac{\pi }{4}+\frac{2}{3}$$

(50)