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!
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It looks to me that the integral tends to a finite value, while the $n + 5$ factor increases, so you get something that grows quite fast. – vonbrand Feb 25 '13 at 20:51
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What have you done so far? – Julien Feb 25 '13 at 20:52
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@vonbrand The integral tends to $0$. It is indeed finite, but actually $n+5$ times the integral is bounded. – Julien Feb 25 '13 at 20:52
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Sorry, misread the upper index. Too little coffee... – vonbrand Feb 25 '13 at 20:58
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@vonbrand The upper bound is indeed not so far from being $3n$ modulo a character... – Julien Feb 25 '13 at 21:07
3 Answers
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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.
Julien
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Hey! thank you very much, it was very clear. I have only one doubt, How do i get that $$\int_0^{3/n}e^{-2x^2}dx\sim \frac{3}{n}.$$?? – user63534 Feb 25 '13 at 22:27
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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.
Sasha
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