calculate Σ_(n=1) ^∞ (((−1)^n )/n) by using ψ (digamma)


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Question Number 174118 by Mathspace last updated on 25/Jul/22

$c a l c u l a t e \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{n} b y u s i n g$ $ψ (d i g a m m a)$
	Answered by mnjuly1970 last updated on 25/Jul/22
	$Σ_(n=1) ^∞ (((−1)^( n) )/n) = Σ_(n=1) ^∞ (1/(2n)) −Σ_(n=1) ^∞ (( 1)/(2n−1)) = (1/2){Σ_(n=1) ^∞ (1/n) −Σ_(n=1) ^∞ (1/(n−(1/2)))} = (1/2) lim_(n→∞) {γ−γ +Σ_(k=1) ^n ((1/k)−(1/(k−(1/2))))} = (γ/2) +(1/2) ψ ((1/2)) = (γ/2)+ (1/2)(−γ−2ln(2)) = −ln(2) ■$
	$\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n} = \underset{n = 1}{\sum^{\infty}} \frac{1}{2 n} - \underset{n = 1}{\sum^{\infty}} \frac{1}{2 n - 1}$ $= \frac{1}{2} {\underset{n = 1}{\sum^{\infty}} \frac{1}{n} - \underset{n = 1}{\sum^{\infty}} \frac{1}{n - \frac{1}{2}}}$ $= \frac{1}{2} {l i m}_{n \to \infty} {γ - γ + \underset{k = 1}{\sum^{n}} (\frac{1}{k} - \frac{1}{k - \frac{1}{2}})}$ $= \frac{γ}{2} + \frac{1}{2} ψ (\frac{1}{2}) = \frac{γ}{2} + \frac{1}{2} (- γ - 2 l n (2))$ $= - l n (2) ◼$
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