TITLE: Moving a limit inside an infinite sum
QUESTION [12 upvotes]: Is uniform convergence justification for moving a limit inside an infinite sum?
I'm trying to evaluate $$ \lim_{n \to \infty} n \int_{0}^{1} \ln(1+x^{n}) \ dx . $$ 
I found that it equals $$\displaystyle\lim_{n \to \infty} \sum_{k=1}^{\infty} (-1)^{k+1} \frac{n}{k+k^2n}.$$
Can I move the limit inside the infinite sum and conclude
$$ \lim_{n \to \infty} n \int_{0}^{1} \ln(1+x^{n}) \ dx = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{2}} = \frac{\pi^{2}}{12}?$$

REPLY [3 votes]: You could also let $x^n = y$, which gives you
$$
n \int_0^1 \log(1+x^n)\,dx = \int_0^1 \frac{\log(1+y)}{y}\,y^{1/n}\,dy \longrightarrow \int_0^1 \frac{\log(1+y)}{y}\,dy,
$$
by the dominated convergence theorem.