# $\lim_{n\to\infty}\int_0^n\left(1-\frac{x}{n}\right)^n\text{e}^{\frac{x}{2}}\text{d}x$ Evaluating this limit

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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Do you know the dominated convergence theorem? – Ayman Hourieh Feb 17 '13 at 2:47
@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
@AntonioVargas : Thank you for letting me know, I try something else and test it too. – Arjang Feb 27 '13 at 1:49
@AntonioVargas : Although ctrl + click would work when there is no test, I'll add some text to avoid that problem, thanks again – Arjang Feb 27 '13 at 1:51
@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? – Arjang Feb 27 '13 at 2:00

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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I see, I have to look over that theorem now:) – Ryan Feb 17 '13 at 3:07
Is it possible to do it in other ways? ie. squeeze theorem? – Ryan Feb 17 '13 at 3:13

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 $n \to \infty$.

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Many thanks! :) – Ryan Feb 17 '13 at 9:36
@Anonio Vargas, What leads you to a upper limit of $n^{\frac14}$? Just observation? :) – Ryan Feb 18 '13 at 0:47
@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 1:18
(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 1:20
Thx for your explanation:) – Ryan Feb 18 '13 at 4:45