.... advanced calculus ... prove that:::: Σ_(n=1) ^∞ {((ζ(2n+1))/(4^(n ) (2n+1)))}=ln(2)−γ γ:: euler−mascheroni constant


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Question Number 123764 by mnjuly1970 last updated on 28/Nov/20
$.... advanced calculus ... prove that:::: Σ_(n=1) ^∞ {((ζ(2n+1))/(4^(n ) (2n+1)))}=ln(2)−γ γ:: euler−mascheroni constant$
$. . . . a d v a n c e d c a l c u l u s . . .$ $p r o v e t h a t ::::$ $\underset{n = 1}{\sum^{\infty}} {\frac{ζ (2 n + 1)}{4^{n} (2 n + 1)}} = l n (2) - γ$ $γ :: e u l e r - m a s c h e r o n i$ $c o n s t a n t$
	Answered by Dwaipayan Shikari last updated on 28/Nov/20

	$\underset{n = 1}{\sum^{\infty}} \frac{ζ (2 n + 1)}{4^{n} (2 n + 1)}$ $\underset{n = 1}{\sum^{\infty}} \underset{k = 1}{\sum^{\infty}} \frac{1}{4^{n} (2 n + 1) k^{(2 n + 1)}}$ $\underset{k = 1}{\sum^{\infty}} \underset{n ⩾ 1}{\sum^{\infty}} \frac{1}{k {(4 k^{2})}^{n} (2 n + 1)}$ $\underset{k ⩾ 1}{\sum^{\infty}} \underset{n ⩾ 1}{\sum^{\infty}} \int_{0}^{1} \frac{x^{2 n}}{k {(4 k^{2})}^{n}} d x$ $\underset{k ⩾ 1}{\sum^{\infty}} \int_{0}^{1} \underset{n ⩾ 1}{\sum^{\infty}} \frac{x^{2 n}}{k {(4 k^{2})}^{n}} d x = \underset{k ⩾ 1}{\sum^{\infty}} \int_{0}^{1} \frac{1}{k} . \frac{\frac{x^{2}}{4 k^{2}}}{1 - \frac{x^{2}}{4 k^{2}}} d x$ $= \underset{k ⩾ 1}{\sum^{\infty}} \int_{0}^{1} \frac{1}{k} . \frac{x^{2}}{4 k^{2} - x^{2}} d x$ $\frac{1}{2} \int_{0}^{1} x (\underset{k ⩾ 1}{\sum^{\infty}} \frac{1}{k} . \frac{1}{(2 k - x)} - \frac{1}{k} . \frac{1}{(2 k + x)}) d x$ $= - \frac{1}{2} (\int_{0}^{1} \underset{k ⩾ 1}{\sum^{\infty}} \frac{1}{k} - \frac{1}{2 k - x} - \int_{0}^{1} \frac{1}{2 k} - \frac{1}{2 k + x})$ $= - \frac{1}{2} ((\int_{0}^{1} \sum^{\infty} \frac{1}{k} - \frac{1}{k - \frac{x}{2}} + \sum^{\infty} \frac{1}{k} - \frac{1}{k + \frac{x}{2}})) d x$ $= - γ - \frac{1}{2} \int_{0}^{1} ψ (- \frac{x}{2}) + ψ (\frac{x}{2})$ $= - γ - [l o g {(Γ (\frac{x}{2}) - l o g (Γ (- \frac{x}{2}))]}_{0}^{1}$ $= - γ + l o g (2)$ $= l o g (2) - γ$
	Commented by mnjuly1970 last updated on 28/Nov/20

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