Measure theory convergence question. 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.
 A: We want to apply the dominated convergence theorem, but as you noticed, we have to check that the function we have to integrate can be bounded by a function which doesn't depend on $n$. $k$ is fixed, so put $f_n(x)=\frac n{(1+\sqrt x)^n}+\frac{n^kx^k}{(1+\sqrt x)^n}$. Since $(1+\sqrt x)^n\geq 1+n\sqrt x$, we have for $x\neq 0$
$$\frac n{(1+\sqrt x)^n}\leq \frac n{1+n\sqrt x}\leq \frac 1{\sqrt x},$$
which is integrable. Since for $k\leq n$ we have $(1+\sqrt x)^n\geq \sum_{l=0}^k\binom nl\sqrt x^l$, and so 
\begin{align*}
\frac{n^kx^k}{(1+\sqrt x)^n}&\leq \frac{n^kx^k}{\sum_{l=0}^k\binom nl\sqrt x^l}\\
&\leq \frac{n^kx^k}{\binom nk\sqrt x^k}\\
&=k!\frac{n^k}{n(n-1)\cdots (n-k+1)}x^{\frac k2}, 
\end{align*} 
and we have for $n$ large enough, say $\geq n_0$ we get $\frac{n^k}{n(n-1)\cdots (n-k+1)}\leq 2$, so for $n\geq n_0$
$$|f_n(x)| \le\frac 1{\sqrt x}+2k!\cdot x^{\frac k2},$$
which is integrable.
A: 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 $O(1/n)$.
