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Question Number 151636 by Tawa11 last updated on 22/Aug/21

Commented by mr W last updated on 22/Aug/21
$do you know the psi−function and ψ(1+z)=−γ+Σ_(n=1) ^∞ ((1/n)−(1/(n+z))) ? if not, then you should learn this at first. Σ_(n=0) ^∞ (1/((n+α)(n+β))) =(1/(β−α))Σ_(n=0) ^∞ [(1/(n+α))−(1/(n+β))] =(1/(β−α)){Σ_(n=0) ^∞ [(1/(n+α))]−Σ_(n=0) ^∞ [(1/(n+β))]} =(1/(β−α)){Σ_(n=1) ^∞ [(1/(n+(α−1)))]−Σ_(n=1) ^∞ [(1/(n+(β−1)))]} =(1/(β−α)){−γ+Σ_(n=1) ^∞ (1/n)−ψ(1+(α−1))−[−γ+Σ_(n=1) ^∞ (1/n)−ψ(1+(β−1))]} =(1/(β−α)){ψ(β)−ψ(α)}$
$d o y o u k n o w t h e p s i - f u n c t i o n a n d$ $ψ (1 + z) = - γ + \underset{n = 1}{\sum^{\infty}} (\frac{1}{n} - \frac{1}{n + z}) ?$ $i f n o t, t h e n y o u s h o u l d l e a r n t h i s a t$ $f i r s t .$ $\underset{n = 0}{\sum^{\infty}} \frac{1}{(n + α) (n + β)}$ $= \frac{1}{β - α} \underset{n = 0}{\sum^{\infty}} [\frac{1}{n + α} - \frac{1}{n + β}]$ $= \frac{1}{β - α} {\underset{n = 0}{\sum^{\infty}} [\frac{1}{n + α}] - \underset{n = 0}{\sum^{\infty}} [\frac{1}{n + β}]}$ $= \frac{1}{β - α} {\underset{n = 1}{\sum^{\infty}} [\frac{1}{n + (α - 1)}] - \underset{n = 1}{\sum^{\infty}} [\frac{1}{n + (β - 1)}]}$ $= \frac{1}{β - α} {- γ + \underset{n = 1}{\sum^{\infty}} \frac{1}{n} - ψ (1 + (α - 1)) - [- γ + \underset{n = 1}{\sum^{\infty}} \frac{1}{n} - ψ (1 + (β - 1))]}$ $= \frac{1}{β - α} {ψ (β) - ψ (α)}$
Commented by Tawa11 last updated on 22/Aug/21

$Please help .$ $Explain how \underset{n = 0}{\sum^{\infty}} \frac{1}{(n + \frac{2}{3}) (n + \frac{4}{3})} = \frac{ψ (\frac{4}{3}) - ψ (\frac{2}{3})}{\frac{4}{3} - \frac{2}{3}}$ $I need the rule please . Explain . Thanks .$
Commented by Tawa11 last updated on 22/Aug/21

$Wow . This is interesting . God bless you sir .$ $Am reading on it . So, I ask the question so that I can read further .$ $Thanks sir . I appreciate .$
Commented by Tawa11 last updated on 22/Aug/21

$After going through your explanation sir . I understand it perfectly .$ $I am happy . God bless you sir$
	Answered by Olaf_Thorendsen last updated on 22/Aug/21

	$S = \frac{2}{9} \underset{n = 0}{\sum^{\infty}} \frac{1}{(n + \frac{2}{3}) (n + \frac{4}{3})}$ $S = \frac{2}{9} . \frac{ψ (\frac{4}{3}) - ψ (\frac{2}{3})}{\frac{4}{3} - \frac{2}{3}}$ $∙ ψ (z + 1) = ψ (z) + \frac{1}{z}$ $\Rightarrow ψ (\frac{1}{3} + 1) = ψ (\frac{1}{3}) + \frac{1}{1 / 3}$ $ψ (\frac{4}{3}) = ψ (\frac{1}{3}) + 3 (1)$ $∙ ψ (1 - z) - ψ (z) = π \cot (π z)$ $ψ (1 - \frac{1}{3}) - ψ (\frac{1}{3}) = π \cot (\frac{π}{3})$ $ψ (\frac{2}{3}) = ψ (\frac{1}{3}) + \frac{π}{\sqrt{3}} (2)$ $∙ (1) - (2) :$ $ψ (\frac{4}{3}) - ψ (\frac{2}{3}) = 3 - \frac{π}{\sqrt{3}}$ $∙ S = \frac{2}{9} . \frac{3 - \frac{π}{\sqrt{3}}}{\frac{2}{3}} = 1 - \frac{π}{3 \sqrt{3}}$
	Commented by Tawa11 last updated on 22/Aug/21

	$Wow . Thanks sir . I appreciate . God bless you .$ $I am visiting all your solution in the forum on this . Thank you sir .$
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