Limit (n to infinity) of a sum where coefficients depend on n. When can I separate the limits?

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InputQuestionOutputConfusion

So I'm trying to evaluate the limit of a sum of the following type: $$ \lim\limits_{n\rightarrow\infty}\sum\limits_{k=1}^na_{k,n} $$ where the $a_{k,n}$ are some coefficients that depend on the running index $k$ but also on $n$. I can't figure out how to derive the limit, but I know the limits of the coefficients, i.e., $$\lim\limits_{n\rightarrow\infty}a_{k,n}=a_k.$$ My question is, when can I do the following: $$ \lim\limits_{n\rightarrow\infty}\sum\limits_{k=1}^na_{k,n}=\lim\limits_{n\rightarrow\infty}\sum\limits_{k=1}^n\lim\limits_{n\rightarrow\infty}a_{k,n}=\lim\limits_{n\rightarrow\infty}\sum\limits_{k=1}^na_k. $$ I was trying to fit the usual interchange of limits / sums rules to this situation but couldn't figure out how they would work here. Assuming that the sum $\sum a_k$ converges I also checked the following: $$ \left|\sum\limits_{k=1}^\infty a_k-\sum\limits_{k=1}^n a_{k,n}\right|\leq\left|\sum\limits_{k=1}^\infty a_k-\sum\limits_{k=1}^n a_k\right|+\sum\limits_{k=1}^n|a_k-a_{k,n}|\leq\varepsilon+n\varepsilon_n $$ where $\varepsilon$ comes from the convergence of $\sum a_k$ and $\varepsilon_n$ comes from the convergence of all $\lim\limits_{n\rightarrow\infty}a_{k,n}$. So for the right-hand-side to go to $0$ I need all sequences to converge faster than $\frac{1}{n}$ I guess?

Does anybody know which Theorem or result would allow taking the limits of the sum and sequences seperately instead of in one go? That would be much easier for the problem I'm facing.

Thank you!
 

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