Limit of an Integral, Then taking Sum 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?
 A: $$\begin{align}\lim_{n\to\infty}\left(\sum_{k=1}^{n}\frac{I_{k+1}}{k}\right)
&= \lim_{n\to\infty} \int_0^1 \left( \sum_{k=1}^n \frac{x^{k+1}}{k} \right) e^x \, \mathrm{d}x
\\&= \lim_{n \to \infty} \int_0^1 \left( -x \log(1-x) + d_{n+1}(x) \right) e^x \,\mathrm{d}x
\\&= \int_0^1 -x \log(1-x) e^x \, \mathrm{d}x
+ \lim_{n \to \infty} \int_0^1 d_{n+1}(x) e^x \mathrm{d}x
\\&= \int_0^1 -x \log(1-x) e^x \, \mathrm{d}x
\end{align}$$
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.
A: Yes, you are right about the first step. 
First of all, let us find the recurrent relation for the $I_n$. 
Integrating by parts we get:
$$I_n=\int^1_0x^ne^x\,dx=e-\int^1_0nx^{n-1}e^x\,dx=e-nI_{n-1}$$
This is equivalent to $I_n+nI_{n-1}=e$. The last equality implies $I_{n+1}+(n+1)I_n=I_n+nI_{n-1}$ or $I_{n+1}=n(I_{n-1}-I_n)$. Now lets analyze our sum. 
We obtain:
$$\sum_{k=1}^n\frac{I_{k+1}}{k}=\sum_{k=2}^n(I_{k-1}-I_k)+I_2=I_1-I_n+I_2$$
Now we must find to which number does $I_n$ tend. 
It's quite simple to note that:
$$0\leq \int^1_0x^ne^x\,dx \leq e\int^1_0x^n\,dx\Leftrightarrow 0\leq I_n\leq \frac{e}{n+1}$$
So we get that:
$$\lim_{n\to\infty}I_n=0$$
To finish our problem, we must integrate by parts to calculate that:
$$\lim_{n\to\infty} \left(I_1+I_2-I_n\right)=I_1+I_2=e-1$$
