How to evaluate : $$\lim_{n\to\infty}\int_0^n\left(1-\frac{x}{n}\right)^n\text{e}^{\frac{x}{2}}\text{d}x$$
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3Do you know the dominated convergence theorem? – Ayman Hourieh Feb 17 '13 at 02:47
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2@Arjang, if a title is only latex then you can't open it with right click -> new tab. Don't take the words out of titles. – Antonio Vargas Feb 26 '13 at 23:47
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@AntonioVargas : Thank you for letting me know, I try something else and test it too. – jimjim Feb 27 '13 at 01:49
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@AntonioVargas : Although ctrl + click would work when there is no test, I'll add some text to avoid that problem, thanks again – jimjim Feb 27 '13 at 01:51
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@AntonioVargas : Antonio I am trying to make the titles shorter so the related side links can fit and are their relevance is clearly visible. Having "Evaluate" vs "How to evaluate" be a help in that regards, but if you know of any problems that would cause before I it instead? – jimjim Feb 27 '13 at 02:00
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@Arjang that's a great point. The best compromise I can think of would be to put some text after the latex, like "$\lim \int n ,dx$ - Evaluating this limit". I agree that there are other options, like ctrl+click or middle click when using a 3-button mouse, but I think we should be going for maximal usability. – Antonio Vargas Feb 27 '13 at 03:30
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@AntonioVargas : load and clear, WILCO – jimjim Feb 27 '13 at 04:08
2 Answers
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Use the fact that $$\left(1 - \dfrac{x}n \right)^n \leq e^{-x}$$ i.e. $$\left(1 - \dfrac{x}n \right)^n e^{x/2} \leq e^{-x/2}$$ and $$\lim_{n \to \infty}\left(1 - \dfrac{x}n \right)^n = e^{-x}$$ Consider the sequence $$f_n(x) = \begin{cases} \left(1 - \dfrac{x}{n} \right)^n e^{x/2} & x \in [0,n]\\ 0 & x > n\end{cases}$$ which is dominated by $g(x) = e^{-x/2}$. Now apply dominated convergence theorem to get that $$\lim_{n \to \infty} \int_0^n f_n(x) dx = \lim_{n \to \infty} \int_0^{\infty} f_n(x) dx = \int_0^{\infty} \lim_{n \to \infty} f_n(x) dx = \int_0^{\infty} e^{-x/2} dx = 2$$
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Here's a proof using just the squeeze theorem (as requested by Ryan in the comments on Marvis' answer).
For $x = o(n)$ and $n$ large enough we have
$$ n \log\left(1-\frac{x}{n}\right) = -\sum_{k=1}^{\infty} \frac{x^k}{k n^{k-1}} \geq - x - \frac{x^2}{n}, $$
so that
$$ \left(1-\frac{x}{n}\right)^n \geq e^{-x-x^2/n}. $$
This gives
$$ \begin{align*} \int_0^n \left(1-\frac{x}{n}\right)^n e^{x/2} \,dx &\geq \int_0^{n^{1/4}} \left(1-\frac{x}{n}\right)^n e^{x/2}\,dx \\ &\geq \int_0^{n^{1/4}} e^{-x/2-x^2/n}\,dx \\ &\geq e^{-1/\sqrt{n}} \int_0^{n^{1/4}} e^{-x/2}\,dx. \end{align*} $$
The last expression converges to
$$ \int_0^\infty e^{-x/2}\,dx = 2 $$ as $n \to \infty$. Marvis' answer shows that
$$ \int_0^n \left(1-\frac{x}{n}\right)^n e^{x/2} \,dx \leq \int_0^\infty e^{-x/2}\,dx = 2, $$
so we conclude from the squeeze theorem that
$$ \int_0^n \left(1-\frac{x}{n}\right)^n e^{x/2}\,dx \longrightarrow 2 $$
as $n \to \infty$.
Antonio Vargas
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@Anonio Vargas, What leads you to a upper limit of $n^{\frac14}$? Just observation? :) – Ryan Feb 18 '13 at 00:47
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1@Ryan There are three main conditions I needed to satisfy. I wanted to replace $n$ by something, say $a_n$, so that $$\begin{align}& a_n \to \infty, \&a_n/n \to 0, \text{ and} \&a_n^2/n \to 0\end{align}$$ as $n \to \infty$. The first condition makes sure the upper limit of integration goes to $\infty$. The second condition makes sure that $x/n \to 0$ for all $x$ in the interval of integration $[0,a_n]$, which is the condition I need to satisfy so that the first inequality in my post is true. These first two requirements are satisfied by $a_n = \sqrt{n}$, for example. (cont.) – Antonio Vargas Feb 18 '13 at 01:18
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1(cont.) Now for $0 \leq x \leq a_n$ we have $e^{-x^2/n} \geq e^{-a_n^2/n}$, and I want this to have a limit of $1$ as $n \to \infty$. This is not true for $a_n = \sqrt{n}$ but it is true for smaller $a_n$, for example $a_n = n^{1/4}$. – Antonio Vargas Feb 18 '13 at 01:20
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