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Question:

Find this limit $$\lim_{n\to\infty}\left(\sqrt{n}\int_{0}^{1}(e^x(1-x))^ndx\right)$$

my idea: since $$e^{x}=\sum_{k=0}^{\infty}\frac{x^k}{k!}$$ so $$(1-x)e^x=\sum_{k=0}^{\infty}\dfrac{(1-x)x^k}{k!}$$ so $$\lim_{n\to\infty}\left(\sqrt{n}\int_{0}^{1}\sum_{k=0}^{\infty}\dfrac{x^k(1-x)}{k!}dx\right)$$ then I fell very ugly

math110
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    Where is the $n$ exponent gone ? –  Sep 15 '14 at 09:23
  • Why do I have the nagging impression that all this "my idea" stuff is mainly here to decorate the post and to pretend that this is not the n-th PSQ by this user? Because I am evil or because it is so? – Did Sep 15 '14 at 10:05

1 Answers1

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It is better to exploit $$\log(1-x) = -x-\frac{x^2}{2}+O(x^3)$$ in order to have: $$\begin{eqnarray*}\int_{0}^{1}((1-x)e^x)^n\,dx &=& \int_{0}^{1}\exp\left(-\frac{n x^2}{2}+n\,O(x^3)\right)\,dx\\ &=&\frac{1}{\sqrt{n}}\int_{0}^{\sqrt{n}}\exp\left(-\frac{x^2}{2}+\frac{O(x^3)}{\sqrt{n}}\right)\,dx\end{eqnarray*}$$ hence the dominated convergence theorem easily gives: $$\lim_{n\to +\infty}\left(\sqrt{n}\int_{0}^{1}((1-x)e^x)^n\,dx\right)=\int_{0}^{+\infty}e^{-x^2/2}\,dx = \color{red}{\sqrt{\frac{\pi}{2}}}.$$

Jack D'Aurizio
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  • Not so sure ! Cheers :-) – Claude Leibovici Sep 15 '14 at 09:44
  • Of course, this is the correct result but, to get a proof in the classical meaning of the term, one should control the approximation in the $O(\ )$ part. – Did Sep 15 '14 at 10:07
  • @Did: To achieve a good control, it is sufficient to prove that over $(0,1)$ we have $$(1-x^3)\leq (1-x)e^{x+x^2/2}\leq(1-x^3/3)$$ – Jack D'Aurizio Sep 15 '14 at 11:49
  • Maybe this is sufficient, if it is, you could show that this holds and how it implies the result. (For the upper bound, try the simpler $1$...) – Did Sep 15 '14 at 12:07