6
$\begingroup$

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 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.

  • 1
    You could try using the fact that $e^x = \lim_{n\rightarrow \infty} (1+x/n)^n$ instead of doing a substitution.2012-10-28

2 Answers 2

5

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.

  • 0
    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.2012-10-30
2

\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*}