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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?
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.
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.
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
$$LHS = \lim_{n\to\infty}\sum_{k=0}^n 0 = \lim_{n\to\infty} 0 = 0$$
$$RHS = \sum_{k=0}^\infty 0 = 0$$
