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I'm trying to teach myself some analysis (I'm currently studying algebra), and I'm a bit stuck on this question. It's strange because of the $n$ appearing as a limit of integration; I want to apply something like LDCT (I guess), but it doesn't seem that can be done directly.

I have noticed that the change of variables $u=1+\frac{x}{2n}$ helps. With this, the problem becomes $$ \lim_{n\to\infty}\int_1^{3/2}2nu^ne^{-2n(u-1)}\,du. $$
This at least solves the issue of the integration limits. Let's let $f_n(u):=2nu^ne^{-2n(u-1)}$ for brevity. I believe it can be shown that $$ \lim_{n\to\infty}f_n(u)=\cases{\infty,\,u=1\\0,\,1<u\leq 3/2} $$ using L'Hopital's rule and the fact that $u^n$ intersects $e^{2n(u-1)}$ where $u=1$, and so the exponential function is larger than $u^n$ for $n>1$.

I think I was also able to show that $\{f_n\}$ is eventually decreasing on $(1,3/2]$, and so Dini's Theorem says that the sequence is uniformly convergent to $0$ on $[u_0,3/2]$ for any $u_0\in (1,3/2]$. Since each $f_n$ is continuous on the closed and bounded interval $[u_0,3/2]$, each is bounded; as the convergence is uniform, the sequence is uniformly bounded.

Thus, the Lebesgue Dominated Convergence Theorem says $$ \lim_{n\to\infty}\int_{u_0}^{3/2}2nu^ne^{-2n(u-1)}\,du=\int_{u_0}^{3/2}0\,du=0. $$ So it looks like I'm almost there, I just need to extend the lower limit all the way to $1$. I think this amounts to asking whether we can switch the order of the limits in $$\lim_{n\to\infty}\lim_{u_0\to 1^+}\int_{u_0}^{3/2}2nu^ne^{-2n(u-1)}\,du, $$ and (finally!) this is where I'm stuck. I feel like this step should be easy, and it's quite possible I'm missing something obvious. That happens a lot when I try to do analysis because of my practically nonexistent background.

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You could try using the fact that $e^x = \lim_{n\rightarrow \infty} (1+x/n)^n$ instead of doing a substitution. – Jair Taylor Oct 28 '12 at 5:33
up vote 5 down vote accepted

HINT Note that $$\left(1 + \dfrac{x}{2n} \right)^n < e^{x/2}$$ for all $n$. Hence, $$ \left(1 + \dfrac{x}{2n} \right)^n e^{-x} < e^{-x/2}$$ Your sequence $$f_n(x) = \begin{cases} \left(1 + \dfrac{x}{2n} \right)^n e^{-x} & x \in [0,n]\\ 0 & x > n\end{cases}$$ is dominated by $g(x) = e^{-x/2}$. Now apply LDCT.

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So, you're saying to look at this problem as $\lim_{n\to\infty}\int_0^\infty f_n(x)\,dx$, where $f_n$ is as you describe? I guess that makes sense... I hadn't thought of looking at it that way, but I guess it should be the same thing. The pointwise limit of the sequence would be $e^{-x/2}$ I suppose, and that would make the answer 2? – topspin1617 Oct 28 '12 at 5:47
@topspin1617 Yes precisely. The answer would be $2$. – user17762 Oct 28 '12 at 6:00
Awesome! Thank you very much. I could definitely see from the beginning that that factor of the integrand approached $e^{x/2}$, but didn't know what to do with the $n$ as the upper limit of the integral. After seeing this answer it seems kind of obvious I guess; I just got stuck on substitution to make the limits constant. I suspect this technique will be useful for many problems. – topspin1617 Oct 30 '12 at 13:18

\begin{eqnarray*} \lim_{n \to \infty}\int_{0}^{n}\left(1+\frac{x}{2n}\right)^n{\rm e}^{-x}\,{\rm d}x & = & \lim_{n \to \infty}n\int_{0}^{1}\left(1+\frac{x}{2}\right)^n{\rm e}^{-nx}\,{\rm d}x = \lim_{n \to \infty}n\int_{0}^{1}{\rm e}^{n\ln\left(1\ +\ x/2\right)-nx}\,{\rm d}x \\[3mm] & = & \lim_{n \to \infty}n\int_{0}^{1}{\rm e}^{-n\,x\,/\,2}\,{\rm d}x = \lim_{n \to \infty}n \left({\rm e}^{-n\,/\,2} - 1 \over -n/2\right) = {\Large 2} \end{eqnarray*}

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