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?
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?
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0@J.M.: Done. Enjoy! – 2011-10-25
2 Answers
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.
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0This is quite a bit beyond what was expected of Zev, but I'll leave it here anyway: *Mathematica* evaluates Zev's starting integral as a Hurwitz zeta function, while both series in the OP evaluate as polygamma functions. The connection between those two functions is well-known to those who know them ;), but the connection would not have been seen without some massaging. – 2011-10-25
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}. $
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0I was expecting you'd be using a as the base and then connect rising and falling factorials, but this is nice! – 2011-10-26