# Integration using Lebesgue dominated convergence theorem

This is an old comp question I'm working on.

$$\lim_{n\to\infty}\int_{[0,1]}\frac{d\lambda}{x^\frac{1}{n}(1+\frac{x}{n})^n}$$

I am having trouble finding a dominating function. Thinking about the blow up when x gets close to zero is is giving me the most trouble. Once I can justify pushing the limit in getting the solution is not a problem. Do I need to use the almost everywhere convergence of $f_n$ to take care of this?

• $x^{1/n}\ge x^{1/2}$ for $n\ge 2$ and $0<x\le 1$ and (trivially) $(1+x/n)^n\ge 1$, so $g(x)=x^{-1/2}$ will work fine as a dominating function. By DC, the limit thus equals $\int_0^1 e^{-x}\, dx=1-e^{-1}$. – user138530 Jul 9 '14 at 1:17
• @ChristianRemling Why don't you turn your comment into an answer? – user940 Jul 9 '14 at 1:20
• @ByronSchmuland: You're welcome to go ahead and answer yourself if you prefer to have an "official" answer rather than a comment. – user138530 Jul 9 '14 at 1:23
• Thanks. I need to work a bunch of these type problems. Setting it up for $n>2$ since were dealing with limits helps. – coffeebelly Jul 9 '14 at 1:30
• @ChristianRemling: Please see this: meta.math.stackexchange.com/questions/1559/…. There are advantages in having an "official" answer (rather than a comment). – Aryabhata Jul 9 '14 at 1:37

Since $x^{1/n}\geq x^{1/2}$ for $n\geq 2$ and $0<x\leq 1$ and (trivially) $(1+x/n)^n \geq 1$ for $n\geq 1$, so $g(x)=x^{-1/2}$ will work fine as a dominating function. By Dominated convergence theorem, the limit thus equals $\int_0^1 e^{-x}\,dx=1-e^{-1}$.