...nice ... ... ... calculus... prove:: Ω=Σ_(n=1) ^∞ ((ζ(2n+1)−1)/(n+1)) =−γ+log(2)


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Question Number 138065 by mnjuly1970 last updated on 09/Apr/21

$. . . n i c e . . . . . . . . . c a l c u l u s . . .$ $p r o v e ::$ $Ω = \underset{n = 1}{\sum^{\infty}} \frac{ζ (2 n + 1) - 1}{n + 1} = - γ + l o g (2)$
	Answered by Ñï= last updated on 09/Apr/21

	$Ω = \underset{k = 2}{\sum^{\infty}} \underset{n = 1}{\sum^{\infty}} \frac{1}{(n + 1) k^{2 n + 1}} = - \underset{k = 2}{\sum^{\infty}} (\frac{1}{k} + l n (1 - \frac{1}{k^{2}}))$ $= \lim_{m \to \infty} (- H_{m} + 1 - \underset{k = 2}{\sum^{\infty}} (k l n (k + 1) + k l n (k - 1) - 2 k l n k))$ $= \lim_{m \to \infty} (- H_{m} + 1 + l n 2 + l n m - m l n (1 + \frac{1}{m}))$ $= \lim_{m \to \infty} (- H_{m} + l n m + l n 2 + O (\frac{1}{m}))$ $= - γ + l n 2$ $. . . . . . i d o n t u n d e r s t a n d . . . j u s t c o p y . . .$
	Commented by mnjuly1970 last updated on 10/Apr/21

	Commented by mnjuly1970 last updated on 10/Apr/21

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