Another New Sequence Which Converges Faster Towards to the Euler-Mascheroni Constant

1. Introduction

It is well known (see [1,2,3,4,5,6]) that the sequence

$${\gamma }_n=1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n}}-lnn,n\ge 1,$$

is convergent to a limit denoted $\gamma =0,5772\dots$ now known as Euler-Mascheroni constant. Many authors have obtained different estimations for ${\gamma }_n-\gamma ;$ for example, the following inequalities increase better [1,2,3,4,5,6]:

$${\displaystyle \frac{1}{2(n+1)}}<{\gamma }_n-\gamma <{\displaystyle \frac{1}{2(n-1)}},n\ge 2,$$

$${\displaystyle \frac{1}{2(n+1)}}<{\gamma }_n-\gamma <{\displaystyle \frac{1}{2n}},n\ge 1,$$

$${\displaystyle \frac{1-\gamma }{n}}<{\gamma }_n-\gamma <{\displaystyle \frac{1}{2n}},n\ge 1,$$

$${\displaystyle \frac{1}{2n+1}}<{\gamma }_n-\gamma <{\displaystyle \frac{1}{2n}},n\ge 1,$$

$${\displaystyle \frac{1}{2n+\frac{2}{5}}}<{\gamma }_n-\gamma <{\displaystyle \frac{1}{2n+\frac{1}{3}}},n\ge 1,$$

$${\displaystyle \frac{1}{2n+\frac{2\gamma -1}{1-\gamma }}}<{\gamma }_n-\gamma <{\displaystyle \frac{1}{2n+\frac{1}{3}}},n\ge 1.$$

The convergence of the sequence ${\gamma }_n$ to $\gamma$ is very slow.

DeTemple [7] modified the logarithmic term of ${\gamma }_n$ and showed that the sequence

$$R_n=1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n}}-ln(n+{\displaystyle \frac{1}{2}})$$

converges to $\gamma$ with rate of convergence $n^{-2}$, since

$${\displaystyle \frac{1}{24{(n+1)}^2}}<R_n-\gamma <{\displaystyle \frac{1}{24n^2}},n\ge 1.$$

Chen [8] proved that for all integers $n\ge 1$,

$${\displaystyle \frac{1}{24{(n+a)}^2}}\le R_n-\gamma <{\displaystyle \frac{1}{24{(n+b)}^2}},$$

with the best possible constants

$$a={\displaystyle \frac{1}{\sqrt{24[-\gamma +1-ln(\frac{3}{2})]}}}-1=0,55106\dots \mathrm{and}b={\displaystyle \frac{1}{2}}.$$

Vernescu [9] provided the sequence

$$V_n=1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n-1}}+{\displaystyle \frac{1}{2n}}-lnn(={\gamma }_n-{\displaystyle \frac{1}{2n}}),$$

which also converges to $\gamma$ with rate of convergence $n^{-2}$, since

$${\displaystyle \frac{1}{12{(n+1)}^2}}<\gamma -V_n<{\displaystyle \frac{1}{12n^2}}.$$

A similar convergence result to $\gamma$ with rate of convergence $n^{-2}$ was obtained by Ivan [10]:

$$li{m}_{n\to \infty }{n}^2(c_n-\gamma )={\displaystyle \frac{1}{6}},$$

where $c_n=1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n}}-ln\sqrt{n(n+1)}$.

Cristea and Mortici [11] introduced the family of sequences

$$v_n(a,b)=1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n-2}}+{\displaystyle \frac{an+b}{n(n-1)}}-lnn,$$

where $a,b$ are real parameters.

Furthermore, they proved that, among the sequences ${(v_n(a,b))}_{n\ge 1}$, the privileged one ${(v_n({\displaystyle \frac{3}{2}},-{\displaystyle \frac{5}{12}}))}_{n\ge 1}$ offers the best approximation to $\gamma$, since it has the rate of convergence $n^{-3}$. More precisely, they obtained the bounds

$${\displaystyle \frac{1}{12n^3}}+{\displaystyle \frac{11}{120n^4}}<v_n({\displaystyle \frac{3}{2}},-{\displaystyle \frac{5}{12}})-\gamma <{\displaystyle \frac{1}{12n^3}}+{\displaystyle \frac{13}{120n^4}}(n\ge 9),$$

where

$$v_n({\displaystyle \frac{3}{2}},-{\displaystyle \frac{5}{12}})=1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n-2}}+{\displaystyle \frac{13}{12(n-1)}}+{\displaystyle \frac{5}{12n}}-lnn.$$

Lu [12] used continued fraction approximation to obtain the following faster sequences converging to the Euler–Mascheroni constant:

$$r_n^{(2)}=1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n}}-{\displaystyle \frac{3}{6n+1}}-lnn,$$

$$r_n^{(3)}=1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n}}-{\displaystyle \frac{6n-1}{12n^2}}-lnn,$$

which satisfie

$${\displaystyle \frac{1}{72{(n+1)}^3}}<\gamma -r_n^{(2)}<{\displaystyle \frac{1}{72n^3}},$$

$${\displaystyle \frac{1}{120{(n+1)}^4}}<r_n^{(3)}-\gamma <{\displaystyle \frac{1}{120{(n-1)}^4}}.$$

Negoi [13] modified the logarithmic term of ${\gamma }_n$ and showed that the sequence

$$T_n=1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n}}-ln(n+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{24n}}),$$

is strictly increasing and convergent to $\gamma$ with rate of convergence $n^{-3}$. Moreover, he proved that

$${\displaystyle \frac{1}{48{(n+1)}^3}}<\gamma -T_n<{\displaystyle \frac{1}{48n^3}},n\ge 1.$$

Chen and Mortici [14] proved that for all integers $n\ge 1$,

$${\displaystyle \frac{1}{48{(n+a)}^3}}\le \gamma -T_n<{\displaystyle \frac{1}{48{(n+b)}^3}},$$

with the best possible constants

$$a={\displaystyle \frac{1}{\sqrt[3]{48[1-\gamma +ln(\frac{37}{24})]}}}-1=0,27380525\dots andb={\displaystyle \frac{83}{360}}=0,23055555\dots .$$

Recently You and Chen [15] modified the logarithmic term of ${\gamma }_n$ and they showed that the sequences

$$r_2(n)=1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n}}-ln(n+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{24n}}+{\displaystyle \frac{1}{48n^2}}),$$

$$r_3(n)=1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n}}-ln(n+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{24n}}-{\displaystyle \frac{1}{48n^2}}-{\displaystyle \frac{1}{48n^3}}-{\displaystyle \frac{1}{576n^4}}-{\displaystyle \frac{1}{1152n^5}}),$$

converge to $\gamma$ with rates of convergence $n^{-3}$, respectively $n^{-4}$, since

$${\displaystyle \frac{1}{24{(n+1)}^3}}<\gamma -r_2(n)<{\displaystyle \frac{1}{24n^3}},$$

$${\displaystyle \frac{143}{5760{(n+1)}^4}}<r_3(n)-\gamma <{\displaystyle \frac{143}{5760n^4}}.$$

By modifying the logarithmic term of ${\gamma }_n$, Cringanu [16] defined a new sequence

$$S_n=1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n}}-ln(n+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{24n}}-{\displaystyle \frac{1}{48n^2}})$$

and showed as he converges to $\gamma$ with rate of convergence $n^{-4}$, since

$${\displaystyle \frac{23}{5760{(n+a)}^4}}\le S_n-\gamma <{\displaystyle \frac{23}{5760{(n+b)}^4}},$$

with the best possible constants $a=\sqrt[4]{{\displaystyle \frac{23}{5760[1-\gamma -ln(\frac{73}{48})]}}}-1=0,0315\dots ,b=-{\displaystyle \frac{7}{46}}=-0,1521\dots$.

Now we define the sequence $L_n=1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n}}-ln(n+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{24n}}-{\displaystyle \frac{1}{48n^2}}+{\displaystyle \frac{23}{5760n^3}})$, for $n\ge 1$,

and we prove that for all integers $n\ge 1$,

$${\displaystyle \frac{17}{3840{(n+\alpha )}^5}}\le L_n-\gamma <{\displaystyle \frac{17}{3840{(n+\beta )}^5}},\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h} \mathrm{t}\mathrm{h}\mathrm{e} \mathrm{b}\mathrm{e}\mathrm{s}\mathrm{t} \mathrm{p}\mathrm{o}\mathrm{s}\mathrm{s}\mathrm{i}\mathrm{b}\mathrm{l}\mathrm{e} \mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{s}$$

$\alpha =\sqrt[5]{{\displaystyle \frac{17}{3840[1-\gamma -ln(\frac{8783}{5760})]}}}-1=0,37407\dots ,\beta ={\displaystyle \frac{3305}{12852}}=0,25715\dots$.

2. The Main Result

Starting from the sequences $R_n$, $T_n$ and $S_n$ we consider the family of sequences

$$L_n(a,b,c,d)=1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n}}-ln(n+a+{\displaystyle \frac{b}{n}}+{\displaystyle \frac{c}{n^2}}+{\displaystyle \frac{d}{n^3}}),$$

for $a,b,c,d\in R,n\ge 1$, and

$$K_n(a,b,c,d)=L_n-\gamma =1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n}}-ln(n+a+{\displaystyle \frac{b}{n}}+{\displaystyle \frac{c}{n^2}}+{\displaystyle \frac{d}{n^3}})-\gamma ,$$

which converges to zero.

Using a Maclaurin growth series we get

$$K_{n+1}(a,b,c,d)-K_n(a,b,c,d)={\displaystyle \frac{1}{n^2}}(a-{\displaystyle \frac{1}{2}})+{\displaystyle \frac{1}{n^3}}(-a^2-a+2b+{\displaystyle \frac{2}{3}})+{\displaystyle \frac{1}{n^4}}(a^3-3ab+{\displaystyle \frac{3a^2}{2}}+a-3b+3c-{\displaystyle \frac{3}{4}})+$$

$$+{\displaystyle \frac{1}{n^5}}(-a^4-2a^3+4a^{2}b-2a^2-2b^2+6ab-4ac-a+4b-6c+4d+{\displaystyle \frac{4}{5}})+$$

$$+{\displaystyle \frac{1}{n^6}}(a^5+{\displaystyle \frac{5a^4}{2}}-5a^{3}b-10a^{2}b+5a{b}^2+5a^{2}c+{\displaystyle \frac{10a^3}{3}}+$$

$${\displaystyle \frac{5a^2}{2}}+5b^2-10ab+10ac-5ad-5bc+a-5b+10c-10d-{\displaystyle \frac{5}{6}})+O({\displaystyle \frac{1}{n^7}}).$$

If $a-{\displaystyle \frac{1}{2}}=0,-a^2-a+2b+{\displaystyle \frac{2}{3}}=0,a^3-3ab+{\displaystyle \frac{3a^2}{2}}+a-3b+3c-{\displaystyle \frac{3}{4}}=0,$

$-a^4-2a^3+4a^{2}b-2a^2-2b^2+6ab-4ac-a+4b-6c+4d+{\displaystyle \frac{4}{5}}=0$ then

$a={\displaystyle \frac{1}{2}},b={\displaystyle \frac{1}{24}},c=-{\displaystyle \frac{1}{48}},d={\displaystyle \frac{23}{5760}}$ and so

$$a^5+{\displaystyle \frac{5a^4}{2}}-5a^{3}b-10a^{2}b+5a{b}^2+5a^{2}c+{\displaystyle \frac{10a^3}{3}}+{\displaystyle \frac{5a^2}{2}}+5b^2-10ab+10ac-5ad-5bc+a-5b+$$

$$10c-10d-{\displaystyle \frac{5}{6}}=-{\displaystyle \frac{17}{768}}.$$

It results that

$$L_n=L_n({\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{24}},-{\displaystyle \frac{1}{48}},{\displaystyle \frac{23}{5760}})=1+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}+\dots +{\displaystyle \frac{1}{n}}-ln(n+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{24n}}-{\displaystyle \frac{1}{48n^2}}+{\displaystyle \frac{23}{5760n^3}})$$

and

$$K_n=K_n({\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{24}},-{\displaystyle \frac{1}{48}},{\displaystyle \frac{23}{5760}})=1+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}+\dots +{\displaystyle \frac{1}{n}}-ln(n+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{24n}}-{\displaystyle \frac{1}{48n^2}}+{\displaystyle \frac{23}{5760n^3}})-\gamma .$$

Thus, we get

$$K_{n+1}-K_n=-{\displaystyle \frac{17}{768n^6}}+O({\displaystyle \frac{1}{n^7}}).$$

By a standard result, if a sequence $x_n$ converges to zero and there exists the ${lim}_{n\to \infty }{n}^k(x_n-x_{n+1})=l$, then ${lim}_{n\to \infty }{n}^{k-1}{x}_n={\displaystyle \frac{l}{k-1}}$ (see e.g., [17]).

In our case of $K_n$, we have ${lim}_{n\to \infty }{n}^6(K_n-K_{n+1})={\displaystyle \frac{17}{768}}$ and so

$$\underset{n\to \infty }{lim}{n}^5{K}_n={\displaystyle \frac{l}{k-1}}={\displaystyle \frac{17}{3840}}.$$

Starting from this result and using an elementary sequence method and MATLAB software for computation, we obtain the following:

Theorem 1.  $Foreveryintegern\ge 1wehave$

$${\displaystyle \frac{17}{3840{(n+\alpha )}^5}}\le L_n-\gamma <{\displaystyle \frac{17}{3840{(n+\beta )}^5}},\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h} \mathrm{t}\mathrm{h}\mathrm{e} \mathrm{b}\mathrm{e}\mathrm{s}\mathrm{t} \mathrm{p}\mathrm{o}\mathrm{s}\mathrm{s}\mathrm{i}\mathrm{b}\mathrm{l}\mathrm{e} \mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{s}$$

$\alpha =\sqrt[5]{{\displaystyle \frac{17}{3840[1-\gamma -ln(\frac{8783}{5760})]}}}-1=0,37407\dots ,\beta ={\displaystyle \frac{3305}{12852}}=0,25715\dots$.

Proof. 

We define the sequence

$$a_n=L_n-\gamma -{\displaystyle \frac{17}{3840{(n+a)}^5}}=1+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}+\dots +{\displaystyle \frac{1}{n}}-ln(n+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{24n}}-{\displaystyle \frac{1}{48n^2}}+{\displaystyle \frac{23}{5760n^3}})-\gamma -{\displaystyle \frac{17}{3840{(n+a)}^5}},$$

for $a>0$ and so $a_{n+1}-a_n=f(n),$ where

$$f(n)={\displaystyle \frac{1}{n+1}}-ln(n+{\displaystyle \frac{3}{2}}+{\displaystyle \frac{1}{24(n+1)}}-{\displaystyle \frac{1}{48{(n+1)}^2}}+{\displaystyle \frac{23}{5760{(n+1)}^3}})+ln(n+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{24n}}-{\displaystyle \frac{1}{48n^2}}+{\displaystyle \frac{23}{5760n^3}})-$$

$$-{\displaystyle \frac{17}{3840{(n+a+1)}^5}}+{\displaystyle \frac{17}{3840{(n+a)}^5}}.$$

The derivative of function f is equal to

$$f^{\prime }(n)=-{\displaystyle \frac{1}{{(n+1)}^2}}-{\displaystyle \frac{3(1920n^4+7680n^3+11440n^2+7600n+1897)}{(n+1)(5760n^4+25920n^3+43440n^2+32040n+8783)}}+$$

$$+{\displaystyle \frac{3(1920n^4-80n^2+80n-23)}{n(5760n^4+2880n^3+240n^2-120n+23)}}-{\displaystyle \frac{17}{768}}{\displaystyle \frac{{(n+a+1)}^6-{(n+a)}^6}{{(n+a)}^6{(n+a+1)}^6}}={\displaystyle \frac{P(n)}{Q(n)}},$$

where

$$P(n)=768(4406400n^4+7490400n^3+2117280n^2-1647796n-606027){(n+a)}^6{(n+a+1)}^6-$$

$$-17n{(n+1)}^2(5760n^4+2880n^3+240n^2-120n+23)(5760n^4+25920n^3+43440n^2+32040n+8783)·$$

$$·[{(n+a+1)}^6-{(n+a)}^6],$$

and

$$Q(n)=768n{(n+1)}^2(5760n^4+2880n^3+240n^2-120n+23)·$$

$$·(5760n^4+25920n^3+43440n^2+32040n+8783){(n+a)}^6{(n+a+1)}^6.$$

By using MATLAB software, we obtain that

$$P(n)=(23688806400a-6091776000)n^{15}+$$

$$+(189510451200a^2+140097945600a-54101237760)n^{14}+$$

$$+(710664192000a^3+1208781619200a^2+283562311680a-216840281088)n^{13}+$$

$$+(1658216448000a^4+4345122816000a^3+3195768176640a^2+27217096704a-517434349824)n^{12}+$$

$$+(2676835123200a^5+9421194240000a^4+11191051468800a^3+$$

$$+4313542551552a^2-951258977280a-816556190976)n^{11}+$$

$$+(3126922444800a^6+13913159270400a^5+22443335884800a^4+$$

$$+15431215165440a^3+2584804713984a^2-2085448478976a-895178414400)n^{10}+$$

$$+(2680219238400a^7+14696194867200a^6+29955583918080a^5+28487564835840a^4+$$

$$+11558431964160a^3-677704596480a^2-2380540040640a-696766207680)n^9+$$

$$+(1675137024000a^8+11256628838400a^7+28108735119360a^6+33662464303104a^5+$$

$$+19582377488640a^4+3364432450560a^3-2380953893760a^2-1694043099360a-386195711526)n^8+$$

$$+(744505344000a^9+6197824512000a^8+18769457479680a^7+27144229367808a^6+19496312727552a^5+$$

$$+5329924312320a^4-1566049991040a^3-1857543344640a^2-777633703566a-149937998355)n^7+$$

$$+(223351603200a^{10}+2382336000000a^9+87842838528008a^8+15008630759424a^7+$$

$$+12269823390720a^6+3216811753728a^5-1958475850560a^4-1844984923200a^3-$$

$$-744713732028a^2-224024317080a-38882029144)n^6+$$

$$+(40609382400a^{11}+603024998400a^{10}+2763719884800a^9+5475548298240a^8+$$

$$+4600589211648a^7+13519070208a^6-2858407439040a^5-2140061613120a^4-$$

$$-720254785500a^3-155641811586a^2-36388303746a-6000763190)n^5+$$

$$+(3384115200a^{12}+89336217600a^{11}+537755811840a^{10}+1188768215040a^9+$$

$$+700593096960a^8-1318070983680a^7-2664208535040a^6-1999216550880a^5-$$

$$-736245769470a^4-1130981320660a^3-12904066140a^2-2408847810a-423322168)n^4+$$

$$+(5752627200a^{12}+54028615680a^{11}+110086612992a^{10}-161048540160a^9-$$

$$-936009699840a^8-1519292712960a^7-1239385743360a^6-540538312566a^5-$$

$$-116468696355a^4-9205598920a^3-99484425a^2-131085198a-40496924)n^3+$$

$$+(1626071040a^{12}-5429661696a^{11}-89850714624a^{10}-310896161280a^9-$$

$$-517564650240a^8-477213143040a^7-247005305856a^6-66245936460a^5-$$

$$-7106635020a^4-235621700a^3-228228570a^2-111896346a-22083544)n^2-$$

$$-(1265507328a^{12}+13178188800a^{11}+49700906496a^{10}+95124456960a^9+$$

$$+102759782400a^8+63444492288a^7+20813514240a^6+2813177334a^5+$$

$$+51512295a^4+68683060a^3+51512295a^2+20604918a+3434153)n-$$

$$-465428736a^{12}-2792572416a^{11}-6981431040a^{10}-9308574720a^9-$$

$$-6981431040a^8-2792572416a^7-465428736a^6.$$

If $23688806400a-6091776000=0,$ so that $a={\displaystyle \frac{3305}{12852}},$ then

$$P(n)=-{\displaystyle \frac{1978301112320}{357}}{n}^{14}-{\displaystyle \frac{6614222397555712}{127449}}{n}^{13}-{\displaystyle \frac{267750738778833992704}{1228480911}}{n}^{12}-$$

$$-{\displaystyle \frac{6410863378097015826387904}{11841327501129}}{n}^{11}-{\displaystyle \frac{11203462396425132464117874560}{12682061753709159}}{n}^{10}-$$

$$-{\displaystyle \frac{283656463491678607733321331433820}{285232250902672695069}}{n}^9-{\displaystyle \frac{9458940152242502727508996226916173}{11979754537912253192898}}{n}^8-$$

$$-{\displaystyle \frac{31432492089149854440508935518270759137255}{70669386642452959618122419064}}{n}^7-$$

$$-{\displaystyle \frac{80478660328007330472740977151857448170459391}{454121478564402718506054664905264}}{n}^6-$$

$$-{\displaystyle \frac{16214760589885408521174449082511424923648820905}{324242735694983541013323030742358496}}{n}^5-$$

$$-{\displaystyle \frac{1566232538811809753079292526805082299432018352507841}{150018035009469424887716193279628490061312}}{n}^4-$$

$$-{\displaystyle \frac{508047894224609605857698089286221126579528398762921383}{275433112277385864093846930861397907752568832}}{n}^3-$$

$$-{\displaystyle \frac{20244841378617164823764709930798913021994554417695397}{68858278069346466023461732715349476938142208}}{n}^2-$$

$$-{\displaystyle \frac{197879846675513964816443737481433155667954876842757889}{6610394694657260738252326340673549786061651968}}n-$$

$$-{\displaystyle \frac{4683428037913070737204329712156882479692563209390625}{8813859592876347651003101787564733048082202624}}<0,$$

for all $n\ge 1,$ and then f is strictly decreasing.

We have $li{m}_{n\to \infty }f(n)=0$ and then it follows $f(n)>0$ for all $n\ge 1,$ such that ${(a_n)}_{n\ge 1}$ is strictly increasing. Since $(a_n)$ converges to zero it results that $a_n<0$ for all $n\ge 1,$ such that

$$L_n-\gamma <{\displaystyle \frac{17}{3840{(n+\frac{3305}{12852})}^5}}, \mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l} n\ge 1.$$

If $a=\sqrt[5]{{\displaystyle \frac{17}{3840[1-\gamma -ln(\frac{8783}{5760})]}}}-1=0,37407\dots ,$ then $a_1=L_1-\gamma -{\displaystyle \frac{17}{3840{(1+a)}^5}}=0$,

$P(n)>0$ for all $n\ge 2$ and then f is strictly increasing on $[2,\infty ).$

Since $li{m}_{n\to \infty }f(n)=0$ it results that $f(n)<0$ for all $n\ge 2,$ such that ${(a_n)}_{n\ge 2}$ is strictly decreasing.

The sequence $(a_n)$ converges to zero and then it results that $a_n>0$ for all $n\ge 2,$ such that

$${\displaystyle \frac{17}{3840{(n+a)}^5}}\le L_n-\gamma , \mathrm{f}\mathrm{o}\mathrm{r} \mathrm{a}\mathrm{l}\mathrm{l} n\ge 1.$$

□

Remark 1. Let us remark that, if $a>{\displaystyle \frac{3305}{12852}},$ then $23688806400a-6091776000>0$ and then there exists $n_a\ge 1$ such that $P(n)>0$ for all $n\ge n_a$ and then

$${\displaystyle \frac{17}{3840{(n+a)}^5}}<L_n-\gamma <{\displaystyle \frac{17}{3840{(n+\frac{3305}{12852})}^5}},$$

for all $n\ge n_a.$

Remark 2. Returning to the sequences

$$R_n=1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n}}-ln(n+{\displaystyle \frac{1}{2}})$$

$$T_n=1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n}}-ln(n+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{24n}})$$

$$S_n=1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n}}-ln(n+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{24n}}-{\displaystyle \frac{1}{48n^2}}),$$

$$L_n=1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n}}-ln(n+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{24n}}-{\displaystyle \frac{1}{48n^2}}+{\displaystyle \frac{23}{5760n^3}})$$

with rates of convergence $n^{-2}$, $n^{-3},n^{-4}$ and respectively $n^{-5}$, after a few iterations we can observe the faster convergence of the sequence $L_n$ to $\gamma =0,577215664901\dots$ compared to $R_n$, $T_n$ and $S_n$:

$$n{R}_n{T}_n{S}_n{L}_n$$

$$100.5775929968\dots 0.5771962501\dots 0.5772160837\dots 0.5772157035\dots$$

$$110.5775303095\dots 0.5772009829\dots 0.5772159500\dots 0.5772156892\dots$$

$$120.5774820339\dots 0.5772042946\dots 0.5772158656\dots 0.5772156808\dots$$

$$130.5774440696\dots 0.5772066809\dots 0.5772158102\dots 0.5772156756\dots$$

$$140.5774136771\dots 0.5772084436\dots 0.5772157727\dots 0.5772156723\dots$$

$$150.5773889693\dots 0.5772097738\dots 0.5772157465\dots 0.5772156702\dots$$

$$160.5773686123\dots 0.5772107964\dots 0.5772157278\dots 0.5772156687\dots$$

$$170.5773516417\dots 0.5772115954\dots 0.5772157142\dots 0.5772156677\dots$$

$$180.5773373461\dots 0.5772122288\dots 0.5772157040\dots 0.5772156670\dots$$

$$190.5773251915\dots 0.5772127372\dots 0.5772156964\dots 0.5772156665\dots$$

$$200.5773147709\dots 0.5772131501\dots 0.5772156905\dots 0.5772156661\dots$$

$$210.5773057696\dots 0.5772134889\dots 0.5772156859\dots 0.5772156659\dots$$

$$220.5772979410\dots 0.5772137694\dots 0.5772156823\dots 0.5772156657\dots$$

$$230.5772910899\dots 0.5772140037\dots 0.5772156795\dots 0.5772156655\dots$$

$$240.5772850602\dots 0.5772142010\dots 0.5772156772\dots 0.5772156654\dots$$

$$250.5772797255\dots 0.5772143682\dots 0.5772156753\dots 0.5772156653\dots$$

$$260.5772749832\dots 0.5772145109\dots 0.5772156738\dots 0.5772156652\dots$$

$$270.5772707485\dots 0.5772146334\dots 0.5772156725\dots 0.5772156651\dots$$

$$280.5772669516\dots 0.5772147391\dots 0.5772156715\dots 0.5772156651\dots$$

$$290.5772635342\dots 0.5772148309\dots 0.5772156706\dots 0.5772156651\dots$$

$$300.5772604473\dots 0.5772149110\dots 0.5772156699\dots 0.5772156650\dots$$

$$400.5772410648\dots 0.5772153449\dots 0.5772156664\dots 0.5772156649\dots$$

$$500.5772320020\dots 0.5772155005\dots 0.5772156654\dots 0.5772156649\dots$$

3. Conclusion

By modifying the logarithmic term of ${\gamma }_n$ we have constructed another new sequence that converges faster to the Euler-Mascheroni constant, with the convergence rate $n^{-5}$, compared to the sequences in [7,8] with convergence rates $n^{-2}$ or those in [13,14,15] with convergence rates $n^{-3}$, or those in [15,16] with convergence rates $n^{-4}$. Also the idea of constructing the sequence $L_n$ allows the construction of a new sequence with the convergence rate $n^{-6}$, starting from the family of sequences

$$L_n^{(5)}(a,b,c,d,e)=1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n}}-ln(n+a+{\displaystyle \frac{b}{n}}+{\displaystyle \frac{c}{n^2}}+{\displaystyle \frac{d}{n^3}}+{\displaystyle \frac{e}{n^4}}),$$

for $a,b,c,d,e\in \mathbf{R}$,$n\ge 1.$

More generally, starting from the family of sequences

$$L_n^{(k)}(a_1,a_2,\dots ,a_k)=1+{\displaystyle \frac{1}{2}}+\dots +{\displaystyle \frac{1}{n}}-ln(n+a_1+{\displaystyle \frac{a_2}{n}}+\dots +{\displaystyle \frac{a_k}{n^{k-1}}}),$$

for $a_1,a_2,\dots ,a_k\in \mathbf{R}$,$k\ge 5,n\ge 1,$ we find a sequence with a convergence rate $n^{-k-1}$.