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Question Number 83629 by M±th+et£s last updated on 04/Mar/20

s h o w t h a t

\underset{n, k = 0}{\sum^{\infty}} \frac{n! k!}{(n + k + 2)!} = \frac{π^{2}}{6}

Answered by Kamel Kamel last updated on 04/Mar/20

S (x) = \underset{n = 0}{\sum^{+ \infty}} \underset{k = 0}{\sum^{+ \infty}} \frac{n! k!}{(n + k + 2)!} x^{n + k + 2} = \underset{n = 0}{\sum^{+ \infty}} \underset{k = 0}{\sum^{+ \infty}} \frac{Γ (n + 1) Γ (k + 1)}{Γ (n + k + 2) (n + k + 2)} x^{n + k + 2}

S^{'} (x) = \underset{n = 0}{\sum^{+ \infty}} \underset{k = 0}{\sum^{+ \infty}} x^{n + k + 1} \int_{0}^{1} t^{n} {(1 - t)}^{k} d t

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

\frac{1}{(1 - x t) (1 - x + x t)} = \frac{1}{2 - x} (\frac{1}{1 - x t} + \frac{1}{1 - x + x t})

∴ S^{'} (x) = \frac{1}{2 - x} \int_{0}^{1} (\frac{1}{1 - x t} + \frac{1}{1 - x + x t}) x d t = - \frac{2}{2 - x} L n (1 - x)

= - \frac{2}{2 - x} L n (1 - x)

∴ S = - 2 \int_{0}^{1} \frac{L n (1 - x)}{2 - x} d x \overset{t = 2 - x}{=} - 2 \int_{1}^{2} \frac{L n (t - 1)}{t} d t

\overset{z = \frac{1}{t}}{=} - 2 \int_{\frac{1}{2}}^{1} \frac{L n (1 - z) - L n (z)}{z} d z

= 2 ({L i}_{2} (1) - {L i}_{2} (\frac{1}{2})) - {L n}^{2} (2)

{L i}_{2} (1 - x) + {L i}_{2} (x) = \frac{π^{2}}{6} - L n (x) L n (1 - x)

∴ {L i}_{2} (\frac{1}{2}) = \frac{π^{2}}{12} - \frac{1}{2} {L n}^{2} (2)

S o : S = 2 (\frac{π^{2}}{6} - \frac{π^{2}}{12} + \frac{1}{2} {L n}^{2} (2)) - {L n}^{2} (2) = \frac{π^{2}}{6}

Commented by M±th+et£s last updated on 04/Mar/20

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