When is the limit of the finite sums equal to the infinite sum of the limits? This question is essentially the same as mine, but the answer and comments only addressed the specific problem there.

Let $f(n,k)$ be a function from $\Bbb N\times\Bbb N$ to $\Bbb R$ such that $\lim\limits_{n\to\infty}f(n,k)$ exists for all $k$ and call it $a_k.$ 
When is it that $$\lim_{n\to\infty}\sum_{k=0}^n f(n,k)= \sum_{k=0}^\infty a_k?$$ I would like to know equivalent or at least sufficient conditions. How does the uniform convergence of $\sum_k f(n,k)$ come into play?
 A: Interchanging limits with integration (or summation) is a central theme in real analysis. Here are two sufficient conditions, following from two well known results:


*

*If $0\le f(n,k)\le f(n+1,k)$ for all $n,k$, then equality holds. This follows from the Monotone Convergence Theorem. Furthermore, the weaker condition $0\le f(n,k)\le a_k$ for all $n,k$ is also sufficient, but this is less well known.

*Let $b_{k}=\sup_{n\ge0}|f(n,k)|$. If $\sum_{k=0}^\infty b_k<\infty$, then equality holds. This follows from the Dominated Convergence Theorem.
A: I would encourage you to think about this in the following way: For all $m,n\in\mathbb{N},$ let $A_{m,n}=\sum_{k=0}^{m}f(n,k).$ You've supposed that $\lim_{n\rightarrow\infty}A_{m,n}=\sum_{k=0}^{m}a_{k}=:A_{m}$ (since we're only adding finitely many terms, we can interchange sum and limit), and the question is whether $\lim_{n\rightarrow\infty}A_{n,n}=\lim_{n\rightarrow\infty}A_{n}=:A_{\infty},$ which we are supposing to exist. I like this notation for the problem because it decouples the summation index ($m$) from the limit index inside the sum ($n$), which can make it easier to think about.
Moreover, given $A_{m,n}$, if $f(n,0)=A_{0,n}$ for all $n,$ and $f(n,k)=A_{k,n}-A_{k-1,n}$ for all $k\geq1$ and all $n,$ then $A_{m,n}=\sum_{k=0}^{m}f(n,k),$ so this is completely equivalent.
If $A_{m,n}$ converges uniformly to $A_{m}$ as $n\rightarrow\infty,$ then given $\varepsilon>0,$ we may find $n$ large enough that $|A_{m,n'}-A_{m}|<\varepsilon$ for every $m$ and all $n'\geq n$. Then in particular, this holds for $m=n,$ which gives $|A_{n,n}-A_{n}|<\varepsilon.$ Once $n$ is large enough, $|A_{n}-A_{\infty}|<\varepsilon,$ which gives $|A_{n,n}-A_{\infty}|<2\varepsilon,$ which is sufficient to complete the proof.
A: One sufficient condition: $$f(n,k)=0$$
Then $a_k = \lim\limits_{n\to\infty}f(n,k)=\lim\limits_{n\to\infty}0 = 0$
Then
$$LHS = \lim_{n\to\infty}\sum_{k=0}^n 0 = \lim_{n\to\infty} 0 = 0$$
$$RHS = \sum_{k=0}^\infty 0 = 0$$ 
