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Is uniform convergence justification for bringing a limit inside of an infinite sum?

I was trying to evaluate $\displaystyle \lim_{n \to \infty} n \int_{0}^{1} \ln(1+x^{n}) \ dx $ and found that it equals $\displaystyle\lim_{n \to \infty} \sum_{k=1}^{\infty} (-1)^{k+1} \frac{n}{k+k^2n}$.

Can I bring the limit inside of the sum?

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If you want to motivate moving a limit inside an integral, you might be interested in [Beppo Levi's theorem][2] and [Lebesgue's theorem][1] (also called dominated convergence theorem). Those are usually used to justify such manipulations. [1]: en.wikipedia.org/wiki/Dominated_convergence_theorem [2]: proofwiki.org/wiki/Beppo_Levi's_Theorem –  Daniel Robert-Nicoud Feb 6 '13 at 17:19

2 Answers 2

up vote 2 down vote accepted

Yes. Uniform convergence allows you to pass the limit inside the sum and inside the integral. The Theorem goes as follows:

Let $X\subseteq \mathbb{R}$, $a\in \bar{\mathbb{R}}$ be a limit point of $X$ (*), $f_k:X\to \mathbb{R}$. If the limit $\lim_{x\to a}f_k(x)$ exists and $\sum_{k=0}^{\infty}f_k$ converges uniformly then \begin{equation}\lim_{x\to a}\sum_{k=0}^{\infty}f_k(x)=\sum_{k=0}^{\infty}\lim_{x\to a}f_k(x)\end{equation}

So in your case you need to show $\lim_{n\to \infty}(-1)^{k+1}\frac{n}{k+k^2n}$ for fixed $k$ exists (simple) and $\sum_{k=0}^{\infty}(-1)^{k+1}\frac{n}{k+k^2n}$ converges uniformly for $n$. I suggest the Weierstrass M test:

If $\forall k\in \mathbb{N}\; \sup_{x\in X}\left|f_k(x)\right|\le M_k$ and the series $\sum_{k=0}^{\infty}M_k$ converges then the series $\sum_{k=0}^{\infty}f_k$ converges uniformly.

In our case, for $x\ge 0$ and $k\ge 1$, $$\left|f_k(x)\right|=\frac{x}{k+k^2x}\le \frac1{k^2}:=M_k$$ and so...

(*) $a$ may very well be infinite

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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.

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