evaluate-n-1-n-1-1-n-1-

Question Number 142290 by mnjuly1970 last updated on 29/May/21

e v a l u a t e :

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

Answered by Dwaipayan Shikari last updated on 29/May/21

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

= \underset{k = 2}{\sum^{\infty}} (- l o g (1 - \frac{1}{k}) - \frac{1}{k})

= \lim_{n \to \infty} - l o g (\frac{1}{2} . \frac{2}{3} . \frac{3}{4} \dots \frac{n}{n + 1}) - \underset{k = 1}{\sum^{\infty}} \frac{1}{k} + 1

= l o g (\frac{2}{1} . \frac{3}{2} . \frac{4}{3} \dots . \frac{n}{n - 1} . \frac{n + 1}{n}) - \underset{k = 1}{\sum^{\infty}} \frac{1}{k} + 1

= \lim_{n \to \infty} l o g (n + 1) - \underset{k = 1}{\sum^{\infty}} \frac{1}{k} + 1 = 1 - γ

Commented by mnjuly1970 last updated on 29/May/21

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