switch finite and infinite rows Asume to have a function $f(k,l):\mathbb{N} \times \mathbb{N}\rightarrow \mathbb{R}$ s.t.
$$\sum_{l=1}^\infty f(k,l)=c_k<C \,\,\,\forall k\in\mathbb{N}.$$
Does the following hold for any finite $n$?:
$$
\sum_{k=1}^n\sum_{l=1}^\infty f(k,l)=\sum_{l=1}^\infty\sum_{k=1}^n f(k,l)
$$
 A: We clearly have that the sum on the LHS is just equal to $\sum_{1 \leq k \leq n} c_k.$ To show that your formula holds, we just need to show that the sequence of the partial sums of the series on the RHS converges to this quantity. Let $\varepsilon > 0.$ Since $\sum_{l \geq 1} f(k, l) = c_k,$ there is $n^\varepsilon_k > 0$ such that $\vert \sum_{1 \leq l \leq m} f(k, l) - c_k \vert < \frac\varepsilon{n}$ for all $m \geq n^\varepsilon_k.$ It follows that for all $m \geq \max_{1 \leq k \leq n} n^\varepsilon_k =: n^\varepsilon,$ we have that $$\vert \sum_{1 \leq l \leq m} \sum_{1 \leq k \leq n} f(k, l) - \sum_{1 \leq k \leq n} c_k \vert \leq \sum_{1 \leq k \leq n} \vert \sum_{1 \leq l \leq m} f(k, l) - c_k \vert < \sum_{1 \leq k \leq n} \frac\varepsilon{n} = \varepsilon,$$ so that $\sum_{l \geq 1} \sum_{1 \leq k \leq n} f(k, l) = \sum_{1 \leq k \leq n} c_k.$ I hope this helps. :)
A: Yes this is true:
Define the sequences $(a_m)_m , (b_m)_m \in \mathbb{R}^{\mathbb{N}}$ (so just real valued sequences) by
$$
a_m := \sum_{i=1}^{n} \sum_{j = 1}^{m} f(i,j) \quad \text{ and } \quad b_m := \sum_{j=1}^{m} \sum_{i = 1}^{n} f(i,j).
$$
What you have given me in the question is that $\lim_{m\rightarrow \infty} a_m$ converges, but we do not now this yet for $\lim_{m\rightarrow \infty} b_m$.
For finite sums you can switch summation (if you would like to prove this rigorously then you can use induction), which implies that $a_m = b_m$ for all $m \in \mathbb{N}$.
In other words the sequence $c_m \subset \mathbb{R}$ (or equivalently $c_m \in \mathbb{R}^{\mathbb{N}}$) defined by $c_m = a_m - b_m$ is constant zero.
This implies that $\lim_{m\rightarrow \infty} c_m = 0$.
From this fact we find
$$
0 = \lim_{m\rightarrow \infty} c_m = \lim_{m\rightarrow \infty} (a_m  - b_m) = \lim_{m\rightarrow \infty} a_m -\lim_{m\rightarrow \infty} b_m,
$$
where in the last equality I used the fact that $\lim_{m\rightarrow \infty} a_m$ exists (which implies that $\lim_{m\rightarrow \infty} b_m$ exists) so we can use the sum rule for limits.
Therefore $\lim_{m\rightarrow \infty} b_m  = \lim_{m\rightarrow \infty}\sum_{i=1}^{n} \sum_{j = 1}^{m} f(i,j) = \sum_{i=1}^{n} \sum_{j = 1}^{\infty} f(i,j)$ exists and equals $\lim_{m\rightarrow \infty} a_m $ (convince yourself that this is what you were asking for).
Some general advice: A priori, it is not clear that the summation on the right hand side (i.e. the sequence $b_m$) converges. So before/after asking yourself if equality occurs you could also ask if this infinite sum even exists. What i'm saying is, do not skip the question of existence too quickly. I tried to emphasize this by explicitly stating that the sequence $b_m$ converges and that I did not assume that the sequence $b_m$ converges.
