Limit $a_{n+1}=\frac{(a_n)^2}{6}(n+5)\int_{0}^{3/n}{e^{-2x^2}} \mathrm{d}x$ I need to find the limit when $n$ goes to $\infty$ of
$$a_{n+1}=\frac{(a_n)^2}{6}(n+5)\int_{0}^{3/n}{e^{-2x^2}} \mathrm{d}x, \quad a_{1}=\frac{1}{4}$$
Thanks in advance!
 A: First show by induction that
$$
0<a_n\leq \frac{1}{4}
$$
for all $n$.
Then use that to show that $a_n$ is decreasing.
Since it is also bounded below, $a_n$ converges to $a\in[0,1/4]$.
Now 
$$
\int_0^{3/n}e^{-2x^2}dx\sim \frac{3}{n}.
$$
So, passing to the limit in the induction formula, we get
$$
a=\frac{a^2}{2}.
$$
So the only possibility is $a=0$.
Hence
$$
\lim_{n\rightarrow +\infty} a_n=0.
$$
Let me know if you want me to expand some points.
A: Obviously $a_n > 0$ for all $n \in \mathbb{N}$. Since
$$
    \int_0^{3/n} \exp(-2 x^2) \mathrm{d}x  < \int_0^{3/n} \mathrm{d}x = \frac{3}{n}
$$
We have
$$
    a_{n+1} < \frac{1}{2} a_n^2 \left(1 + \frac{5}{n} \right) \leqslant  3 a_n^2
$$
Consider sequence $b_n$, such that $b_1 = a_1$ and $b_{n+1} = 3 b_n^2$, then $a_n \leqslant b_n$ by induction on $n$. But $b_n$ admits a closed form solution:
$$
      b_n = \frac{1}{3} \left(\frac{3}{4} \right)^{2^{n-1}}
$$
and $\lim_{n \to \infty }b_n = 0$. Thus, since $0 < a_n \leqslant b_n$,  $\lim_{n \to \infty} a_n = 0$ by squeeze theorem.
A: HINT: apply the mean value theorem on the integral.
