Evaluating $\lim _{n \rightarrow \infty} (2n+1) \int_0 ^{1} x^n e^x dx$ 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.
 A: Following the Ishan Banerjee comment let $t=x^n$ hence $dx=\frac{1}{n}t^{\frac{1}{n}-1}dt$ and then
$$ (2n+1) \int_0 ^{1} x^n e^x dx=\frac{2n+1}{n}\int_0^1t^{1/n}e^{t^{1/n}}dt\to2e$$
by using the dominated convergence theorem.
A: Let $I_n=\int\limits_0^1x^n\mathrm e^x\mathrm dx$. By integration by parts, $(n+1)I_n=\left.x^{n+1}\mathrm e^x\right|_0^1-I_{n+1}=\mathrm e-I_{n+1}$. Now, $0\leqslant I_{n+1}\leqslant I_n$ hence $(n+1)I_n\leqslant\mathrm e\leqslant(n+2)I_n$. 
This is enough to show that 
$$
\left(2-\frac3n\right)\cdot\mathrm e\leqslant(2n+1)I_n\leqslant2\mathrm e,
$$
hence
$(2n+1)I_n\to2\mathrm e$.
A: 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 $.
A: 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 
$$\lim_{n\to\infty}(2n+1)\int_0^1x^ne^xdx=2e.$$
