Ω= Σ_(n=1) ^∞ n(ζ (1+n) −1) =?


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Question Number 163704 by mnjuly1970 last updated on 09/Jan/22

$Ω = \underset{n = 1}{\sum^{\infty}} n (ζ (1 + n) - 1) = ?$
	Answered by Kamel last updated on 09/Jan/22

	$Ψ (1 + x) + \frac{1}{1 + x} = 1 - γ - (\underset{n = 1}{\sum^{+ \infty}} (ζ (n + 1) - 1) {(- x)}^{n}$ $∴ \underset{x \to - 1}{l i m} (Ψ^{(1)} (1 + x) - \frac{1}{{(1 + x)}^{2}}) = \underset{n = 1}{\sum^{+ \infty}} n (ζ (n + 1 - 1)$ $Ψ^{(1)} (1 + x) = \underset{k = 0}{\sum^{+ \infty}} \frac{1}{{(k + x + 1)}^{2}} = \frac{1}{{(1 + x)}^{2}} + \underset{k = 1}{\sum^{+ \infty}} \frac{1}{{(1 + x + k)}^{2}}$ $∴ \underset{n = 1}{\sum^{+ \infty}} n (ζ (n + 1) - 1) = \underset{k = 1}{\sum^{+ \infty}} \frac{1}{k^{2}} = \frac{π^{2}}{6}$
	Commented by mnjuly1970 last updated on 10/Jan/22

	$v e r y n i c e s o l u t i o n . . .$
	Answered by qaz last updated on 10/Jan/22

	$\underset{n = 1}{\sum^{\infty}} n (ζ (1 + n) - 1)$ $= \underset{n = 1}{\sum^{\infty}} \underset{k = 2}{\sum^{\infty}} \frac{n}{k^{1 + n}}$ $= \underset{k = 2}{\sum^{\infty}} \underset{n = 0}{\sum^{\infty}} \frac{n + 1}{k^{n + 2}}$ $= \underset{k = 2}{\sum^{\infty}} \frac{1}{k^{2}} (D + 1) ∣_{x = 1 / k} \frac{1}{1 - x}$ $= \underset{k = 2}{\sum^{\infty}} \frac{1}{k^{2}} \cdot \frac{2 - \frac{1}{k}}{{(1 - \frac{1}{k})}^{2}}$ $= \underset{k = 1}{\sum^{\infty}} (\frac{1}{k^{2}} + \frac{1}{k} - \frac{1}{k + 1})$ $= \frac{π^{2}}{6} + 1$
	Commented by mnjuly1970 last updated on 10/Jan/22

	$t h a n k y o u s o m u c h s i r$ $\frac{π^{2}}{6}$
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