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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 \dots

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}