advanced-calculus-prove-that-n-1-2n-1-4-n-2n-1-ln-2-euler-mascheroni-constant-

Question Number 123764 by mnjuly1970 last updated on 28/Nov/20

\dots . a d v a n c e d c a l c u l u s \dots

p r o v e t h a t ::::

\underset{n = 1}{\sum^{\infty}} {\frac{ζ (2 n + 1)}{4^{n} (2 n + 1)}} = l n (2) - γ

γ :: e u l e r - m a s c h e r o n i

c o n s t a n t

Answered by Dwaipayan Shikari last updated on 28/Nov/20

\underset{n = 1}{\sum^{\infty}} \frac{ζ (2 n + 1)}{4^{n} (2 n + 1)}

\underset{n = 1}{\sum^{\infty}} \underset{k = 1}{\sum^{\infty}} \frac{1}{4^{n} (2 n + 1) k^{(2 n + 1)}}

\underset{k = 1}{\sum^{\infty}} \underset{n ⩾ 1}{\sum^{\infty}} \frac{1}{k {(4 k^{2})}^{n} (2 n + 1)}

\underset{k ⩾ 1}{\sum^{\infty}} \underset{n ⩾ 1}{\sum^{\infty}} \int_{0}^{1} \frac{x^{2 n}}{k {(4 k^{2})}^{n}} d x

\underset{k ⩾ 1}{\sum^{\infty}} \int_{0}^{1} \underset{n ⩾ 1}{\sum^{\infty}} \frac{x^{2 n}}{k {(4 k^{2})}^{n}} d x = \underset{k ⩾ 1}{\sum^{\infty}} \int_{0}^{1} \frac{1}{k} . \frac{\frac{x^{2}}{4 k^{2}}}{1 - \frac{x^{2}}{4 k^{2}}} d x

= \underset{k ⩾ 1}{\sum^{\infty}} \int_{0}^{1} \frac{1}{k} . \frac{x^{2}}{4 k^{2} - x^{2}} d x

\frac{1}{2} \int_{0}^{1} x (\underset{k ⩾ 1}{\sum^{\infty}} \frac{1}{k} . \frac{1}{(2 k - x)} - \frac{1}{k} . \frac{1}{(2 k + x)}) d x

= - \frac{1}{2} (\int_{0}^{1} \underset{k ⩾ 1}{\sum^{\infty}} \frac{1}{k} - \frac{1}{2 k - x} - \int_{0}^{1} \frac{1}{2 k} - \frac{1}{2 k + x})

= - \frac{1}{2} ((\int_{0}^{1} \sum^{\infty} \frac{1}{k} - \frac{1}{k - \frac{x}{2}} + \sum^{\infty} \frac{1}{k} - \frac{1}{k + \frac{x}{2}})) d x

= - γ - \frac{1}{2} \int_{0}^{1} ψ (- \frac{x}{2}) + ψ (\frac{x}{2})

= - γ - [l o g {(Γ (\frac{x}{2}) - l o g (Γ (- \frac{x}{2}))]}_{0}^{1}

= - γ + l o g (2)

= l o g (2) - γ

Commented by mnjuly1970 last updated on 28/Nov/20

G o d k e e p y o u

e x c e l l e n t m r p a y a n \dots