calculate-n-1-1-n-1-n-1-1-

Question Number 160928 by mnjuly1970 last updated on 09/Dec/21

c a l c u l a t e

Ω = \underset{n = 1}{\sum^{\infty}} \frac{ζ (1 + n) - 1}{n + 1} \overset{?}{=} 1 - γ

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Answered by qaz last updated on 09/Dec/21

\underset{n = 1}{\sum^{\infty}} \frac{ζ (1 + n) - 1}{n + 1}

= \underset{n = 1}{\sum^{\infty}} \underset{k = 2}{\sum^{\infty}} \frac{1}{k^{1 + n} (1 + n)}

= \underset{k = 2}{\sum^{\infty}} \underset{n = 2}{\sum^{\infty}} \frac{1}{{nk}^{n}}

= \underset{k = 2}{\sum^{\infty}} (\underset{n = 1}{\sum^{\infty}} (\frac{1}{{nk}^{n}}) - \frac{1}{k})

= \underset{k = 2}{\sum^{\infty}} (\ln \frac{k}{k - 1} - \frac{1}{k})

= \lim_{N \to \infty} \underset{k = 1}{\sum^{N}} (\ln (k + 1) - lnk - \frac{1}{k + 1})

= \lim_{N \to \infty} (\ln (N + 1) - H_{N + 1} + 1)

= 1 - γ

Commented by mnjuly1970 last updated on 09/Dec/21

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