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?
Answer
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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