advanced-calculus-prove-that-n-0-n-1-x-1-n-1-n-x-2-x-1-proof-n-0-n-1-x-1-n-1-n-0-

Question Number 137668 by mnjuly1970 last updated on 05/Apr/21

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

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

Ω = \underset{n = 0}{\sum^{\infty}} \frac{Γ (n + 1) Γ (x + 1)}{(n + 1) Γ (n + x + 2)} = ψ^{'} (x + 1)

p r o o f ::

Ω = \underset{n = 0}{\sum^{\infty}} \frac{β (n + 1, x + 1)}{n + 1} = \underset{n = 0}{\sum^{\infty}} {\frac{1}{(n + 1)} \int_{0}^{1} t^{n} . {(1 - t)}^{x} d t}

= \int_{0}^{1} {{(1 - t)}^{x} \underset{n = 0}{\sum^{\infty}} \frac{t^{n}}{n + 1} d t}

= \int_{0}^{1} {{(1 - t)}^{x} \underset{n = 1}{\sum^{\infty}} \frac{t^{n - 1}}{n} d t}

= - \int_{0}^{1} \frac{{(1 - t)}^{x} l n (1 - t)}{t} d t = - \int_{0}^{1} \frac{t^{x} l n (t)}{1 - t} d t

= \frac{\partial}{\partial x} \int_{0}^{1} \frac{1 - t^{x}}{1 - t} d t = ψ^{'} (1 + x) \dots . . ✓ ✓

\dots \dots . Ω = ψ^{'} (1 + x) \dots \dots

Commented by Dwaipayan Shikari last updated on 05/Apr/21

\underset{n = 0}{\sum^{\infty}} \frac{Γ (n + 1) Γ (1)}{(n + 1) Γ (n + 2)} = \underset{n = 0}{\sum^{\infty}} \frac{1}{n^{2}} = \frac{π^{2}}{6} = ψ^{'} (1)

\underset{n = 0}{\sum^{\infty}} \frac{Γ (n + 1) Γ (2)}{(n + 1) Γ (n + 3)} = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} (n + 1)} = \frac{π^{2}}{6} - 1 = ψ^{'} (2)

\dots

Commented by mnjuly1970 last updated on 05/Apr/21

g r a t e f u l . .

t h a n k y o u s o m u c h m r p a y a n \dots