nice-calculus-prove-n-1-2n-1-1-n-1-log-2-

Question Number 138065 by mnjuly1970 last updated on 09/Apr/21

\dots n i c e \dots \dots \dots c a l c u l u s \dots

p r o v e ::

Ω = \underset{n = 1}{\sum^{\infty}} \frac{ζ (2 n + 1) - 1}{n + 1} = - γ + l o g (2)

Answered by Ñï= last updated on 09/Apr/21

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

= \lim_{m \to \infty} (- H_{m} + 1 - \underset{k = 2}{\sum^{\infty}} (k l n (k + 1) + k l n (k - 1) - 2 k l n k))

= \lim_{m \to \infty} (- H_{m} + 1 + l n 2 + l n m - m l n (1 + \frac{1}{m}))

= \lim_{m \to \infty} (- H_{m} + l n m + l n 2 + O (\frac{1}{m}))

= - γ + l n 2

\dots \dots i d o n t u n d e r s t a n d \dots j u s t c o p y \dots

Commented by mnjuly1970 last updated on 10/Apr/21

Commented by mnjuly1970 last updated on 10/Apr/21