Find out A =Σ_(n=2) ^∞ ((ζ(n))/(n(−3)^n )) where ζ(p)=Σ


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Question Number 78490 by ~blr237~ last updated on 18/Jan/20

$Find out A = \underset{n = 2}{\sum^{\infty}} \frac{ζ (n)}{n {(- 3)}^{n}} where ζ (p) = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{p}}$
	Answered by mind is power last updated on 19/Jan/20
	$let f(x)=Σ_(n≥2) ((ζ(n)x^n )/n)=Σ_(n≥2) (x^n /n)Σ_(m≥1) (1/m^n ) =Σ_(n≥2) .Σ_(m≥1) ((x/m))^n .(1/n) ∀n,m ,n≥2 and m≥1 we haveΣ_(n≥2) Σ_(m≥1) ((x/m))^n .(1/n),≤Σ_(n≥2) (x^n /n).ζ(2)<∞ ,∀x∈]−1,1[ ⇒Σ_(n≥2) .Σ_(m≥1) ((x/m))^n .(1/n)=Σ_(m≥1) Σ_(n≥2) ((((x/m))^n )/n)=f(x) Σ_(k≥1) (a^k /k)=−ln(1−a)⇒Σ_(k≥2) (a^k /k)=−ln(1−a)−a f(x)=Σ_(m≥1) {−ln(1−(x/m))−(x/m)}=f(x) we have our sum=f(−(1/3)) Γ(x)=(1/x)Π_(k≥1) (((1+(1/k))^x )/(1+(x/k)))⇒(1/x)Π_(k≥1) .(e^(xln(1+(1/k))) /(1+(x/k)))=(1/x).Π_(k≥1) (e^(xln(1+(1/k))−(x/k)+(x/k)) /(1+(x/k))) =(1/x).Π_(k≥1) (e^(x(ln(1+(1/k))−(1/k))+(x/k)) /(1+(x/k)))=(1/x).e^(Σ_(k≥1) x(ln(1+(1/k))−(1/k))) .Π_(k≥1) (e^(x/k) /e^(ln(1+(x/k))) ) Σ_(k≥1) {ln(1+(1/k))−(1/k)}=−γ Euler macheronie Constent Γ(x)=(e^(−γx) /x).e^(Σ_(k≥1) {(x/k)−ln(1+(x/k))}) ⇒lnΓ(x)=−γx−ln(x)+Σ_(k≥1) {(x/k)−ln(1+(x/k))} ⇒Σ_(k≥1) {(x/k)−ln(1+(x/k))}=ln(Γ(x))+γx+ln(x) ⇒Σ_(k≥1) {−(x/k)−ln(1−(x/k))}=f(x)=ln(Γ(−x))−γx+ln(−x) S=f(−(1/3))=ln(Γ((1/3)))+(γ/3)+ln((1/3))=Σ_(n≥2) ((ζ(n))/(n(−3)^n ))$
	$let f (x) = \sum_{n ⩾ 2} \frac{ζ (n) x^{n}}{n} = \sum_{n ⩾ 2} \frac{x^{n}}{n} \sum_{m ⩾ 1} \frac{1}{m^{n}}$ $= \sum_{n ⩾ 2} . \sum_{m ⩾ 1} {(\frac{x}{m})}^{n} . \frac{1}{n}$ $\forall n, m, n ⩾ 2 and m ⩾ 1$ $we have \sum_{n ⩾ 2} \sum_{m ⩾ 1} {(\frac{x}{m})}^{n} . \frac{1}{n}, ⩽ \sum_{n ⩾ 2} \frac{x^{n}}{n} . ζ (2) < \infty, \forall x \in] - 1, 1 [$ $\Rightarrow \sum_{n ⩾ 2} . \sum_{m ⩾ 1} {(\frac{x}{m})}^{n} . \frac{1}{n} = \sum_{m ⩾ 1} \sum_{n ⩾ 2} \frac{{(\frac{x}{m})}^{n}}{n} = f (x)$ $\sum_{k ⩾ 1} \frac{a^{k}}{k} = - \ln (1 - a) \Rightarrow \sum_{k ⩾ 2} \frac{a^{k}}{k} = - \ln (1 - a) - a$ $f (x) = \sum_{m ⩾ 1} {- \ln (1 - \frac{x}{m}) - \frac{x}{m}} = f (x)$ $we have our sum = f (- \frac{1}{3})$ $Γ (x) = \frac{1}{x} \prod_{k ⩾ 1} \frac{{(1 + \frac{1}{k})}^{x}}{1 + \frac{x}{k}} \Rightarrow \frac{1}{x} \prod_{k ⩾ 1} . \frac{e^{xln (1 + \frac{1}{k})}}{1 + \frac{x}{k}} = \frac{1}{x} . \prod_{k ⩾ 1} \frac{e^{xln (1 + \frac{1}{k}) - \frac{x}{k} + \frac{x}{k}}}{1 + \frac{x}{k}}$ $= \frac{1}{x} . \prod_{k ⩾ 1} \frac{e^{x (\ln (1 + \frac{1}{k}) - \frac{1}{k}) + \frac{x}{k}}}{1 + \frac{x}{k}} = \frac{1}{x} . e^{\sum_{k ⩾ 1} x (\ln (1 + \frac{1}{k}) - \frac{1}{k})} . \prod_{k ⩾ 1} \frac{e^{\frac{x}{k}}}{e^{\ln (1 + \frac{x}{k})}}$ $\sum_{k ⩾ 1} {\ln (1 + \frac{1}{k}) - \frac{1}{k}} = - γ Euler macheronie Constent$ $Γ (x) = \frac{e^{- γ x}}{x} . e^{\sum_{k ⩾ 1} {\frac{x}{k} - \ln (1 + \frac{x}{k})}}$ $\Rightarrow \ln Γ (x) = - γ x - \ln (x) + \sum_{k ⩾ 1} {\frac{x}{k} - \ln (1 + \frac{x}{k})}$ $\Rightarrow \sum_{k ⩾ 1} {\frac{x}{k} - \ln (1 + \frac{x}{k})} = \ln (Γ (x)) + γ x + \ln (x)$ $\Rightarrow \sum_{k ⩾ 1} {- \frac{x}{k} - \ln (1 - \frac{x}{k})} = f (x) = \ln (Γ (- x)) - γ x + \ln (- x)$ $S = f (- \frac{1}{3}) = \ln (Γ (\frac{1}{3})) + \frac{γ}{3} + \ln (\frac{1}{3}) = \sum_{n ⩾ 2} \frac{ζ (n)}{n {(- 3)}^{n}}$
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