TITLE: Evaluating $\lim _{n \rightarrow \infty} (2n+1) \int_0 ^{1} x^n e^x dx$
QUESTION [6 upvotes]: Could you help me evaluate $\lim _{n \rightarrow \infty} (2n+1) \int_0 ^{1} x^n e^x dx$?
I've calculated that the recurrence relation for this integral is:
$\int_0 ^{1} x^n e^x dx = x^ne^x | ^{1} _{0} - n \cdot \int_0 ^{1} x^{n-1} e^x dx$
So if we let $I_n = \int_0 ^{1} x^n e^x \ dx$, we get $I_n = \left.x^ne^x \,\right |^1 _0 - n \cdot I_{n-1}$.
Can this be useful here?
I would appreciate all your help.

REPLY [3 votes]: A related technique. You can use integration by parts technique by letting $u=e^{x}$ which leads to
$$ I_n = \left( 2\,n+1 \right)  \left( {\frac {{{\rm e}}}{n+1}}-{\frac {{
{\rm e}}}{2+3\,n+{n}^{2}}}+\int _{0}^{1}\!{\frac {{x}^{n+2}{
{\rm e}^{x}}}{ \left( n+2 \right)  \left( n+1 \right) }}{dx} \right) .$$
$$\implies \lim_{n\to \infty}I_n = 2 \,\rm{e} + 0 + \lim_{n\to \infty } \int _{0}^{1}\!{\frac {(2n+1){x}^{n+2}{
{\rm e}^{x}}}{ \left( n+2 \right)  \left( n+1 \right) }}{dx} $$
$$ \implies \lim_{ n\to \infty } = 2 \rm e . $$ 
Note that, the change of the limit with integral is due to the uniform convergence of the sequence $$ \frac {(2n+1){x}^{n+2}}{ \left( n+2 \right)  \left( n+1 \right) }. $$
Here is a technique for proving uniform convergence. 
Added: Integration by parts,
$$ \int u\, dv = u\,v -\int v \,du. $$
So, in your case $ u = e^{x} $ and $ dv = x^n dx $.

REPLY [3 votes]: Here is an alternative argument. Note that for every $a\in [0,1)$,
$$\frac{(1-a^{n+1})e^a}{n+1}=e^a\int_a^1x^ndx\le\int_a^1x^ne^xdx\le\int_0^1x^ne^xdx\le e \int_0^1x^ndx=\frac{e}{n+1}.$$
Multiplying the inequality above by $2n+1$ and letting $n\to\infty$, it follows that:
$$2e^a\le\liminf_{n\to\infty}(2n+1)\int_0^1x^ne^xdx\le\limsup_{n\to\infty}(2n+1)\int_0^1x^ne^xdx\le 2e.$$ 
Since $a\in [0,1)$ is arbitrary, we can conclude that 
$$\lim_{n\to\infty}(2n+1)\int_0^1x^ne^xdx=2e.$$