Σ_(n=1) ^∞ (1/(n^3 +n^2 +n+1))


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Question Number 123326 by Dwaipayan Shikari last updated on 24/Nov/20

$\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{3} + n^{2} + n + 1}$
Commented by Dwaipayan Shikari last updated on 25/Nov/20

$I h a v e f o u n d \frac{γ - 1}{2} + \frac{1}{4 i} ((1 - i) ψ (1 - i) - (1 + i) ψ (i) - 1 - i)$
	Answered by mnjuly1970 last updated on 25/Nov/20
	$=Σ(1/((n^2 +1)(n+1)))=Σ_(n=1) ^∞ (1/((n−i)(n+i)(n+1))) =(1/(2i))Σ_(n=1) ^∞ [(1/((n−i)(n+1))) −(1/((n+i)(n+1)))] (1/(2i))Σ_(n=1) ^∞ ((1/((n−i)(n+1))))−(1/(2i))Σ_(n=1) ^∞ ((1/((n+i)(n+1)))) (1/(2i(1+i))){Σ_(n=1) ^∞ (1/(n−i)) −(1/(n+1))}−(1/(2i(1−i))){Σ_(n=1) ^∞ (1/(n+i))−(1/(n+1))} =(1/(2i(1+i)))(ψ(2)−ψ(1−i))−(1/(2i(1−i)))(ψ(2)−ψ(1+i)) =[(1/(2i(1+i)))−(1/(2i(1−i)))]ψ(2)+(1/(2i(1−i)))ψ(1+i)−(1/(2i(1+i)))ψ(1−i) =((−2i)/(2i(1+1)))ψ(2)+{(1/(2i))[((ψ(1+i))/(1−i))−((ψ(1−i))/(1+i))]} =((−1)/2)(1−γ)+{(1/(4i))[(1+i)ψ(1+i)−(1−i)ψ(1−i)]}=A =(1/2)(γ−1)+A^? ✓$
	$= Σ \frac{1}{(n^{2} + 1) (n + 1)} = \underset{n = 1}{\sum^{\infty}} \frac{1}{(n - i) (n + i) (n + 1)}$ $= \frac{1}{2 i} \underset{n = 1}{\sum^{\infty}} [\frac{1}{(n - i) (n + 1)} - \frac{1}{(n + i) (n + 1)}]$ $\frac{1}{2 i} \underset{n = 1}{\sum^{\infty}} (\frac{1}{(n - i) (n + 1)}) - \frac{1}{2 i} \underset{n = 1}{\sum^{\infty}} (\frac{1}{(n + i) (n + 1)})$ $\frac{1}{2 i (1 + i)} {\underset{n = 1}{\sum^{\infty}} \frac{1}{n - i} - \frac{1}{n + 1}} - \frac{1}{2 i (1 - i)} {\underset{n = 1}{\sum^{\infty}} \frac{1}{n + i} - \frac{1}{n + 1}}$ $= \frac{1}{2 i (1 + i)} (ψ (2) - ψ (1 - i)) - \frac{1}{2 i (1 - i)} (ψ (2) - ψ (1 + i))$ $= [\frac{1}{2 i (1 + i)} - \frac{1}{2 i (1 - i)}] ψ (2) + \frac{1}{2 i (1 - i)} ψ (1 + i) - \frac{1}{2 i (1 + i)} ψ (1 - i)$ $= \frac{- 2 i}{2 i (1 + 1)} ψ (2) + {\frac{1}{2 i} [\frac{ψ (1 + i)}{1 - i} - \frac{ψ (1 - i)}{1 + i}]}$ $= \frac{- 1}{2} (1 - γ) + {\frac{1}{4 i} [(1 + i) ψ (1 + i) - (1 - i) ψ (1 - i)]} = A$ $= \frac{1}{2} (γ - 1) + \overset{?}{A} ✓$
	Commented by Dwaipayan Shikari last updated on 25/Nov/20

	$G r e a t! t h a n k i n g y o u s i r$
	Commented by mnjuly1970 last updated on 25/Nov/20

	$p e a c e b e u p o n y o u$ $s i r p a y a n .$
	Commented by mnjuly1970 last updated on 25/Nov/20

	Commented by mnjuly1970 last updated on 25/Nov/20

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