TITLE: Measure theory convergence question.
QUESTION [7 upvotes]: Show $$\lim_{n \to \infty} \int_0^1 \frac{n+n^k x^k}{(1+\sqrt{x})^n}dx=0$$ for $k=1,2,3,...$ It's clear that the functions converge pointwise to $0$ on $(0,1]$ but I can't seem to find an integrable dominating function. Any hints would be much appreciated.

REPLY [2 votes]: Here's a more detailed analysis of your integral. By change of variables, write
$$ \int_0^1 \frac{n+n^k x^k}{(1+\sqrt{x})^n}dx = {1\over n}\int_0^n \frac{2s(1+s^{2k}/n^{k+1})}{(1+s/n)^n}ds$$
Note that for fixed $s$, we have $\displaystyle\frac{2s(1+s^{2k}/n^{k+1})}{(1+s/n)^n}\downarrow 2s\exp(-s)$, and that if $n\geq 2k+3$ then 
$$\int_0^\infty \frac{2s(1+s^{2k}/n^{k+1})}{(1+s/n)^n}ds<\infty.$$
So by dominated convergence, 
$$n \int_0^1 \frac{n+n^k x^k}{(1+\sqrt{x})^n}dx\to \int_0^\infty 2s\exp(-s)\,ds=2.$$
In particular, the original integral is $O(1/n)$.