calculate Σ_(n=2) ^∞ (((−1)^n )/n)ξ(n)


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Question Number 82447 by mathmax by abdo last updated on 21/Feb/20

$c a l c u l a t e \sum_{n = 2}^{\infty} \frac{{(- 1)}^{n}}{n} ξ (n)$
	Answered by mind is power last updated on 21/Feb/20
	$=Σ_(n≥2) (((−1)^n )/n).Σ_(m≥1) (1/m^n ) =Σ_(m≥1) Σ_(n≥2) (((−1)^n )/(nm^n )) ln(1+x)=Σ_(k≥1) (((−1)^(k−1) x^k )/k) ⇒Σ_(n≥1) (((−1)^n )/n).((1/m))^n =−Σ_(n≥1) (((−1)^(n−1) )/n).((1/m))^n =−ln(1+(1/m)) ⇒Σ_(n≥2) (((−1)^n )/n)((1/m))^n =−ln(1+(1/m))+(1/m) Σ_(m≥1) {−ln(1+(1/m))+(1/m)}=lim_(x→∞) {−ln(1+(1/m))+(1/m)} =lim_(x→∞) Σ_(m=1) ^x {(1/m)}−ln(1+x)=γ$
	$= \sum_{n ⩾ 2} \frac{{(- 1)}^{n}}{n} . \sum_{m ⩾ 1} \frac{1}{m^{n}}$ $= \sum_{m ⩾ 1} \sum_{n ⩾ 2} \frac{{(- 1)}^{n}}{{n m}^{n}}$ $l n (1 + x) = \sum_{k ⩾ 1} \frac{{(- 1)}^{k - 1} x^{k}}{k}$ $\Rightarrow \sum_{n ⩾ 1} \frac{{(- 1)}^{n}}{n} . {(\frac{1}{m})}^{n} = - \sum_{n ⩾ 1} \frac{{(- 1)}^{n - 1}}{n} . {(\frac{1}{m})}^{n} = - l n (1 + \frac{1}{m})$ $\Rightarrow \sum_{n ⩾ 2} \frac{{(- 1)}^{n}}{n} {(\frac{1}{m})}^{n} = - l n (1 + \frac{1}{m}) + \frac{1}{m}$ $\sum_{m ⩾ 1} {- l n (1 + \frac{1}{m}) + \frac{1}{m}} = \lim_{x \to \infty} {- l n (1 + \frac{1}{m}) + \frac{1}{m}}$ $= \lim_{x \to \infty} \underset{m = 1}{\sum^{x}} {\frac{1}{m}} - l n (1 + x) = γ$
	Commented by mathmax by abdo last updated on 21/Feb/20

	$t h a n k y o u s i r .$
	Commented by mind is power last updated on 21/Feb/20

	$w i t h e p l e a s u r$
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