TITLE: Limit of an Integral, Then taking Sum
QUESTION [11 upvotes]: I am given that $I_n=\int^1_0x^ne^x\,dx$
Now, how can I find the value of the following limit:
$$\lim_{n\to\infty}\left(\sum_{k=1}^{n}\frac{I_{k+1}}{k}\right)$$
I suppose solving for $I_n$ is that necessary first here?

REPLY [5 votes]: $$\begin{align}\lim_{n\to\infty}\left(\sum_{k=1}^{n}\frac{I_{k+1}}{k}\right)
&= \lim_{n\to\infty} \int_0^1 \left( \sum_{k=1}^n \frac{x^{k+1}}{k} \right) e^x \, \mathrm{d}x
\\&= \lim_{n \to \infty} \int_0^1 \left( -x \log(1-x) + d_{n+1}(x) \right) e^x \,\mathrm{d}x
\\&= \int_0^1 -x \log(1-x) e^x \, \mathrm{d}x
+ \lim_{n \to \infty} \int_0^1 d_{n+1}(x) e^x \mathrm{d}x
\\&= \int_0^1 -x \log(1-x) e^x \, \mathrm{d}x
\end{align}$$
where $d_n(x)$ is the error term for the $n$-th order Taylor series for $-x \log(1-x)$. All that remains is to show that that last integral is $e-1$ (the integrand has an elementary antiderivative) and to show the last limit really does converge to zero; e.g. by bounding the error term. (note that it's an improper integral) (also note that it's not enough to observe $d_{n+1}(x) \to 0$ as $n \to \infty$, because the limit is outside the integral, not inside)
This may not be an easier way, but I did want to show that you didn't have to find $I_{k+1}$ first.