How to prove that $\sum\limits_{n=1}^\infty\frac{(n-1)!}{n\prod\limits_{i=1}^n(a+i)}=\sum\limits_{k=1}^\infty \frac{1}{(a+k)^2}$ for $a>-1$?

A problem on my (last week's) real analysis homework boiled down to proving that, for $a>-1$, $$\sum_{n=1}^\infty\frac{(n-1)!}{n\prod\limits_{i=1}^n(a+i)}=\sum_{k=1}^\infty \frac{1}{(a+k)^2}.$$ Mathematica confirms this is true, but I couldn't even prove the convergence of the original series (the one on the left), much less demonstrate that it equaled this other sum; the ratio test is inconclusive, and the root test and others seem hopeless. It was (and is) quite a frustrating problem. Can someone explain how to go about tackling this?

Solution 1:

This uses a reliable trick with the Beta function. I say reliable because you can use the beta function and switching of the integral and sum to solve many series very quickly.

First notice that $$\prod_{i=1}^{n}(a+i)=\frac{\Gamma(n+a+1)}{\Gamma(a+1)}.$$ Then

$$\frac{(n-1)!}{\prod_{i=1}^{n}(a+i)}=\frac{\Gamma(n)\Gamma(a+1)}{\Gamma(n+a+1)}=\text{B}(n,a+1)=\int_{0}^{1}(1-x)^{n-1}x{}^{a}dx.$$ Hence, upon switching the order we have that $$\sum_{n=1}^{\infty}\frac{(n-1)!}{n\prod_{i=1}^{n}(a+i)}=\int_{0}^{1}x^{a}\left(\sum_{n=1}^{\infty}\frac{(1-x)^{n-1}}{n}\right)dx.$$ Recognizing the power series, this is $$\int_{0}^{1}x^{a}\frac{-\log x}{1-x}dx.$$ Now, expand the power series for $\frac{1}{1-x}$ to get $$\sum_{m=0}^{\infty}-\int_{0}^{1}x^{a+m}\log xdx.$$ It is not difficult to see that $$-\int_{0}^{1}x^{a+m}\log xdx=\frac{1}{(a+m+1)^{2}},$$ so we conclude that $$\sum_{n=1}^{\infty}\frac{(n-1)!}{n\prod_{i=1}^{n}(a+i)}=\sum_{m=1}^{\infty}\frac{1}{(a+m)^{2}}.$$

Hope that helps,

Remark: To evaluate the earlier integral, notice that $$-\int_{0}^{1}x^{r}\log xdx=\int_{1}^{\infty}x^{-(r+2)}\log xdx=\int_{0}^{\infty}e^{-u(r+1)}udu=\frac{1}{(r+1)^{2}}\int_{0}^{\infty}e^{-u}udu. $$ Alternatively, as Joriki pointed out, you can just use integration by parts.

Solution 2:

Since at least J. M. asked for it, here's another solution for the case when $a$ is a natural number.

I'll use the forward difference operator $\Delta$, defined by $\Delta f(n) = f(n+1) - f(n)$, and the falling factorial defined by $$ n^{\underline{a}} = \begin{cases} n(n-1)(n-2) \dots (n-a+1), & a > 0, \\ 1, & a=0 \\ \frac{1}{(n+1)(n+2) \dots (n+|a|)}, & a < 0, \end{cases} $$ and satisfying $\Delta n^{\underline{a}} = a n^{\underline{a-1}}$.

The summand, which I'll denote by $F_a(n)$, can be rewritten as $$ F_a(n) = \frac{(n-1)!}{n\prod_{i=1}^n(a+i)} = \frac{(n-1)! a!}{n (a+n)!} = \frac{a!}{n \cdot n(n+1)(n+2) \dots (n+a)} $$ $$= \frac{(a-1)!}{n} \left( -(-a) (n-1)^{\underline{-(a+1)}}\right) = -\frac{(a-1)!}{n} \Delta\left( (n-1)^{\underline{-a}}\right). $$ Using the rule $\Delta(f(n)g(n)) = \Delta f(n) \, g(n+1) + f(n) \Delta g(n)$, we get $$ F_a(n) = - \Delta\left( \frac{(a-1)!}{n} (n-1)^{\underline{-a}}\right) + \Delta\left( \frac{(a-1)!}{n} \right) \, n^{\underline{-a}} $$ $$ = - \Delta\left( \frac{(a-1)!}{n \cdot n (n+1) \dots (n+a-1)} \right) + (a-1)! \left( \frac{1}{n+1} - \frac{1}{n} \right) \frac{1}{(n+1)\dots (n+a)} $$ $$= - \Delta\left( \frac{(a-1)!}{n \cdot n(n+1) \dots (n+a-1)} \right) + F_{a-1}(n+1) - (a-1)! \Delta\left( \frac{(n-1)^{\underline{-a}}}{-a} \right). $$

Summing over $n \ge 1$ gives (because of telescoping in the sums-of-deltas) $$ \sum_{n=1}^{\infty} F_a(n) = \frac{(a-1)!}{1 \cdot a!} + \sum_{n=1}^{\infty} F_{a-1}(n+1) - \frac{(a-1)!}{a} 0^{-\underline{a}} $$ $$ = \frac{1}{a} + \sum_{m=2}^{\infty} F_{a-1}(m) - \frac{1}{a^2} $$ $$ = \sum_{m=1}^{\infty} F_{a-1}(m) - \frac{1}{a^2} $$ (since $F_{a-1}(1) = 1/a$).

Finally, since $F_0(n) = 1/n^2$, we obtain after using this result to work our way down $n$ steps that $$ \sum_{n=1}^{\infty} F_a(n) = \sum_{n=1}^{\infty} F_{a-1}(n) - \frac{1}{a^2} = \dots = \sum_{n=1}^{\infty} F_0(n) - \left( \frac{1}{a^2} + \dots + \frac{1}{1^2} \right) = \sum_{n=a+1}^{\infty} \frac{1}{n^2} = \sum_{k=1}^{\infty} \frac{1}{(k+a)^2}. $$