If $f_j$ is continuous then $\sum f_j $ is also continuous let $A = \{0,1,1/2,1/3,....\}$. For each $ j \in \mathbb{N}$ define a function $f_j$ on a by $$f_j(0) = \sum_{k=0}^{\infty}\frac{k^j2^j}{k!j!} \hspace{1cm} f_j(\frac{1}{n}) = \sum_{k=0}^{n}\frac{k^j2^j}{k!j!}$$ Then
If $f(x) = \sum_{0}^{\infty}f_j(x), x\in A$. Show that f is continuous at $0$.
My attempt
I could show that that each $f_j$ is well defined using ratio test and thus $f_j$ is continuous at $0$ for each j. But how to proceed from there? any other approach is welcome.
 A: You can interchange the sums because all summands are nonnegative. Now, written as: $f(0)=\sum_k (k!)^{-1}\sum_j (2k)^j/j!$ the inner sum looks like an exponential. It's not hard to see that $\sum_k(k!)^{-1}\sum_j (2k)^{j}/j!=\sum_k\dfrac{e^{2k}}{k!}$. Let $x=e^2$. Then this is our good friend $\sum_k\dfrac{x^k}{k!}=e^{x}=\exp(\exp(2))=f(0)$. Observe that $f(1/n)=\sum_{j}f_{j}(1/n)=\sum_{j}\sum_{k=0}^{n}\dfrac{k^j 2^j}{k! j!}=\sum_{k=0}^{n}\dfrac{e^{2k}}{k!}$.
Let $x=e^2$ as before. Let $S_n(x)$ be the $n$-th partial sum of $\sum_{k=0}^{n}x^k/k!$. Then $S_n(x)=f(1/n)$ and $S_n(x)\to f(0)$ as $n\to\infty$. Sequential continuity is enough here in the subspace topology of $A$ and since $S_n(x)=f(1/n)$ increases with $n$ and is bounded above, it is clear than any sequence in $A$ converging to $0$ converges to $f(0)$.
How to Interchange the Limits
(Proof Sketch)
Let $(x_{n,k})$ be a seq. in $\mathbf{R}_{>0}$ indexed by $(n,k)\in \mathbf{N}\times\mathbf{N}$. Let $(S_{m,j})$ be the seq. in $\mathbf{R}_{>0}$ indexed by $(n,k)\in \mathbf{N}\times\mathbf{N}$ such that $S_{m,j}=\sum_{n=0}^{m}\sum_{k=0}^{j}x_{n,k}$. Then $S_{a,b}\le S_{a+1,b}$, $S_{a,b}\le S_{a,b+1}$ for any $(a,b)\in\mathbf{N}^2$. Let $B=\sup_{(m,j)\in\mathbf{N}^2}S_{m,j}$. If $B=\infty$, we're obviously done. So suppose $B<\infty$. Clearly $B=\lim_{\ell\to\infty}S_{\ell,\ell}$ and also $\lim_{m\to\infty} S_{m,j}\le \lim_{m\to\infty} S_{m,j+1}\le \lim_{m\to\infty} S_{m,j+2}\le\cdots\to B$ and similarly $\lim_{j\to\infty} S_{m,j}\le \lim_{j\to\infty} S_{m+1,j}\le \lim_{j\to\infty} S_{m+2,j}\le\cdots\to B$ (to see this, it helps to recall how $B$ is defined). So you can change the order.
