Is taking the lim of the series the same to taking the series of the lims? [duplicate]

Given non negative real sequences $(a_{1,k})_k, (a_{2,k})_k, ..., (a_{n,k})_k$,... with $n$ all over $\mathbb{N}$. Assume that $\lim\limits_{n \to \infty} a_{n,k}$ exists for all $k$ and $\sum\limits_{k=1}^{\infty} a_{n,k}$ exists for all $n$.

Is it necessary that $\lim\limits_{n \to \infty} \sum\limits_{k=1}^{\infty} a_{n,k} = \sum\limits_{k=1}^{\infty} \lim\limits_{n \to \infty} a_{n,k}$, if both of the sides is either a real number of $+\infty$?

My friends and I were arguing over this, and still we came to no conclusion. If it isn't necessary, in what conditions will the equality holds?

marked as duplicate by Community♦Jul 1 '18 at 4:18

• No, it is not necessary. If the series $\sum_{k=1}^\infty a_{n,k}$ converges uniformly in $n$, then you can exchange limit and summation. The monotone convergence theorem and the dominated convergence theorem also give sufficient conditions. – eloiprime Jul 1 '18 at 3:33
• See the Moore-Osgood Theorem. – Mark Viola Jul 1 '18 at 3:59

Consider

$$a_{nk} = \frac{k}{k+n}-\frac{k-1}{k-1+n}$$

Then

$$\lim_{n \to \infty}\sum_{k=1}^\infty a_{nk} = \lim_{n \to \infty}\lim_{K \to \infty}\sum_{k=1}^K a_{nk} =\lim_{n \to \infty}\lim_{K \to \infty}\frac{K}{K+n} = 1,$$

but

$$\sum_{k=1}^\infty \lim_{n \to \infty}a_{nk} = 0$$

For some conditions where the equality holds, study the monotone and dominated convergence theorems for series.