Is this short proof of Tannery's Theorem (swapping sum and limit operators) correct? I am trying to prepare a proof of Tannery's Theorem that is understandable by non-advanced students. Many of the textbook proofs, and answers on this site (eg here), are overly concise and not accessible.
Is the following correct?

Statement of Theorem
The theorem has three requirements

*

*An infinite sum $S_{j}=\sum_{k}f_{k}(j)$ that converges

*The limit $\lim_{j\rightarrow\infty}f_{k}(j)=f_{k}$ exists

*An $M_{k}\geq\left|f_{k}(j)\right|$ independent of $j$, where $\sum_{k}M_{k}$ converges

If the requirements are met, we can take the limit inside the sum.
$$\lim_{j\rightarrow\infty}\sum_{k}f_{k}(j)=\sum_{k}\lim_{j\rightarrow\infty}f_{k}(j)$$

Proof
Let's first show the sum of the limits actually exists.
By definition, $\left|f_{k}(j)\right|\leq M_{k}$, and $\sum_{k}M_{k}$ converges, so $\sum_{k}\left|f_{k}(j)\right|$ also converges. Therefore $\sum_{k}f_{k}(j)$ converges absolutely, including as $j\rightarrow\infty$. That is, the sum of limits $\sum_{k}\lim_{j\rightarrow\infty}f_{k}(j)$ converges.

Now let's show the limit of the sum is the sum of the limits.
The following easy inequality will be useful.
$$\left|f_{k}(j)-f_{k}\right|\leq\left|f_{k}(j)\right|+\left|f_{k}\right|\leq M_{k}+M_{k}=2M_{k}$$
Since $\sum_{k}M_{k}$ converges there must be an $N$ so that $\sum_{k=N}M_{k}<\epsilon$, where $\epsilon$ is as small as we require.
We can therefore say,
$$\left|\sum_{k=N}f_{k}(j)\right|\leq\sum_{k=N}\left|f_{k}(j)\right|\leq\sum_{k=N}M_{k}<\epsilon$$
The following is the case when $j\rightarrow\infty$.
$$\left|\sum_{k=N}f_{k}\right|\leq\sum_{k=N}\left|f_{k}\right|\leq\sum_{k=N}M_{k}<\epsilon$$

Let's now consider the absolute difference between $\sum_{k}f_{k}(j)$ and $\sum_{k}f_{k}$. Although the following looks complicated, it is simply splitting the sums over $[0,\infty]$ into sums over $[0,N-1]$ and $[N,\infty]$.
$$\begin{align}\left|\sum_{k}f_{k}(j)-\sum_{k}f_{k}\right|&=\left|\sum_{k}^{N-1}f_{k}(j)+\sum_{k=N}f_{k}(j)-\sum_{k}^{N-1}f_{k}-\sum_{k=N}f_{k}\right| \\ \\
&\leq\left|\sum_{k=N}f_{k}(j)\right|+\left|\sum_{k=N}f_{k}\right|+\left|\sum_{k}^{N-1}f_{k}(j)-\sum_{k}^{N-1}f_{k}\right| \\ \\
&<2\epsilon+\left|\sum_{k}^{N-1}\left(f_{k}(j)-f_{k}\right)\right|\end{align}$$
As $j\rightarrow\infty$, the finite sum $\sum_{k}^{N-1}\left(f_{k}(j)-f_{k}\right)\rightarrow0$, which leaves a simpler inequality.
$$\lim_{j\rightarrow\infty}\left|\sum_{k}f_{k}(j)-\sum_{k}f_{k}\right|<2\epsilon$$
Because $\epsilon$ can be as small as we require, we finally have $\lim_{j\rightarrow\infty}\sum_{k}f_{k}(j)=\sum_{k}f_{k}$, which proves the theorem.
$$\lim_{j\rightarrow\infty}\sum_{k}f_{k}(j)=\sum_{k}f_{k}=\sum_{k}\lim_{j\rightarrow\infty}f_{k}(j)$$

 A: The line "Therefore $\sum_{k}f_{k}(j)$ converges absolutely, including
$j\rightarrow\infty$. That is, the sum of limits $\sum_{k}\lim_{j\rightarrow\infty}f_{k}(j)$
converges" needs justification because we cannot take limit and let $j\rightarrow \infty$ within an infinite sum.
Convergence of $\sum_{k}f_{k}$ is easy: Observe that for each $j,k$,
$|f_{k}(j)|\leq M_{k}$. Let $j\rightarrow\infty$, then we get $|f_{k}|\leq M_{k}$.
Since $\sum_{k}M_{k}<\infty$, $\sum_{k}f_{k}$ converges absolutely.
Denote $S(j)=\sum_{k}f_{k}(j)$ and $S=\sum_{k}f_{k}$. To show that
$S(j)\rightarrow S$, we approximate $S(j)$ and $S$ by their partial
sums. This is the well-known $\varepsilon/3$ argument.
Let $\varepsilon>0$ be arbitrary. Choose $K$ such that $\sum_{k>K}M_{k}<\varepsilon$.
(This is possible because $\sum_{k}M_{k}<\infty)$.
For any $j$, we have that $\left|S(j)-\sum_{k=1}^{K}f_{k}(j)\right|\leq\sum_{k\geq K}|f_{k}(j)|\leq\sum_{k\geq K}M_{k}<\varepsilon.$
Similarly, $\left|S-\sum_{k=1}^{K}f_{k}\right|\leq\sum_{k>K}|f_{k}|\leq\sum_{k>K}M_{k}<\varepsilon$.
Then, we choose $J$ such that $|f_{k}(j)-f_{k}|<\frac{\varepsilon}{K}$
whenever $k=1,\ldots,K$ and $j\geq J$. This is possible because
$f_{1}(j)\rightarrow f_{1}$, $f_{2}(j)\rightarrow f_{2}$, ..., $f_{K}(j)\rightarrow f_{K}$
as $j\rightarrow\infty$ and there are finitely many such terms.
Let $j\geq J$ be arbitrary, then
\begin{eqnarray*}
\left|S(j)-S\right| & \leq & \left|S(j)-\sum_{k=1}^{K}f_{k}(j)\right|+\left|\sum_{k=1}^{K}f_{k}(j)-\sum_{k=1}^{K}f_{k}\right|+\left|\sum_{k=1}^{K}f_{k}-S\right|\\
 & \leq & \varepsilon+\sum_{k=1}^{K}|f_{k}(j)-f_{k}|+\varepsilon\\
 & \leq & 2\varepsilon+\sum_{k=1}^{K}\frac{\varepsilon}{K}\\
 & = & 3\varepsilon.
\end{eqnarray*}
This shows that $S(j)\rightarrow S$ as $j\rightarrow\infty$.
