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Evaluating

$$\sum_{m=0}^\infty \sum_{n=0}^\infty\frac{m!n!}{(m+n+2)!}$$

involving binomial coefficients.

My attempt: $$\frac{1}{(m+1)(n+1)}\sum_{m=0}^\infty \sum_{n=0}^\infty\frac{(m+1)!(n+1)!}{(m+n+2)!}=\frac{1}{(m+1)(n+1)} \sum_{m=0}^\infty \sum_{n=0}^\infty\frac{1}{\binom{m+n+2}{m+1}}=?$$

Is there any closed form of this expression?

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    $\begingroup$ How can you take the factor out of the summation ? $\endgroup$
    – Shailesh
    Jan 3, 2016 at 1:57
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    $\begingroup$ @Shailesh. Also,Olivier Oloa. Thanks for pointing out $\endgroup$
    – Maverick
    Jan 3, 2016 at 2:27

2 Answers 2

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One may observe that, $$ \frac{m!}{(m+n+2)!}=\frac{m!}{(n+1)(m+n+1)!}-\frac{(m+1)!}{(n+1)(m+n+2)!} $$ giving, by telescoping terms, $$ \sum_{m=0}^N\frac{m!}{(m+n+2)!}=\frac1{(n+1)(n+1)!}-\frac{(N+1)!}{(n+1)(N+n+2)!} $$ thus, as $N \to \infty$, $$ \sum_{m=0}^\infty\frac{m!}{(m+n+2)!}=\frac1{(n+1)(n+1)!}. $$ Then the initial series reduces to

$$ \sum_{n=0}^\infty \sum_{m=0}^\infty\frac{n!\:m!}{(m+n+2)!}=\sum_{n=0}^\infty\frac{n!}{(n+1)(n+1)!}=\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6. $$

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  • $\begingroup$ Sneaky... :-$)$ $\endgroup$
    – Lucian
    Jan 3, 2016 at 5:27
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    $\begingroup$ Such masterful detelescoping of factorials! (+1) $\endgroup$ Feb 12, 2016 at 16:27
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Another (way less subtle) approach. We have:

$$ S=\sum_{m\geq 0}\sum_{n\geq 0}\frac{\Gamma(m+1)\,\Gamma(n+1)}{(m+n+2)\,\Gamma(m+n+2)}=\sum_{m,n\geq 0}\iint_{(0,1)^2} x^m(1-x)^n y^{m+n+1}\,dx\,dy \tag{1}$$ hence: $$ S = \iint_{(0,1)^2}\frac{y\,dx\,dy}{(1-xy)(1-y+xy)}=2\int_{0}^{1}\frac{-\log(1-y)}{2-y}\,dy=2\int_{0}^{1}\frac{-\log(t)}{1+t}\,dt\tag{2} $$ and by expanding $\frac{1}{1+t}$ as a geometric series, $$ S = 2\sum_{n\geq 0}\frac{(-1)^n}{(n+1)^2} = \color{red}{\zeta(2)} = \frac{\pi^2}{6}.\tag{3}$$

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