Convergence of a series to prove continuity.

Question

Let $A = \{0, 1, \frac{1}{2}, \frac{1}{3}, \cdots\}.$ For each $j \in \mathbb{N}$ define a function $f_j$ on $A$ by $$f_j(0) = \sum_{k=0}^{\infty} \frac{k^j 2^j}{k! j!}, \; f_j(\frac{1}{n}) = \sum_{k=0}^{n} \frac{k^j 2^j}{k! j!}$$

(a) Show that $f_j$ is well-defined and continuous at $0 \in A$ for each $j \in \{0,1,2,3, \cdots\}.$

(b) Let $f(x) = \sum_{j = 0}^{\infty} f_j(x), \; x \in A$. Show that $f$ is continuous at $0$ in $A$.

Attempt:

(a) I take $a_k=\frac{k^j 2^j}{k! j!}$. Then by ratio test $\sum a_k$ converges absolutely. Then, it is bounded. Now if I can show it converges uniformly, then it would be continuous (Since each $a_k$ is continuous) . I believe I have to use the Weierstrass M-Test but what would be the bound M?

(b) We'll have a double sum. Now we have to show that the double sum converges uniformly. By Part(a) $f_j$ converges so again we can bound that by some M. And try to repeat the Weierstrass M-Test. Is this correct?

Answer 1

You only have to consider continuity at $0$ the space $A$ so you don't have to worry about uniform convergence. In a metric space, sequential continuity implies continuity, so for part a) you just have to show that if $x_k\to 0$ then $f_j(x_k)\to f_j(0)$ and that's true by the definition of an infinite sum. once you know it converges.

Now reconsider part b).

For part b) $$\begin{align} f(\frac{1}{n})&= \sum_{j=0}^{\infty}{\sum_{k=0}^n{\frac{k^j2^j}{k!j!}}} =\sum_{k=0}^n{\sum_{j=0}^{\infty}{\frac{k^j2^j}{k!j!}}}\\ \lim_{n\to\infty}f(1/n)&= \sum_{k=0}^{\infty}{\sum_{j=0}^{\infty}{\frac{k^j2^j}{k!j!}}}= \sum_{k=0}^{\infty}{\frac{1}{k!}\sum_{j=0}^{\infty}{\frac{(2k)^j}{j!}}}\\ &=\sum_{k=0}^{\infty}{\frac{e^{2k}}{k!}}= \sum_{k=0}^{\infty}{\frac{(e^2)^k}{k!}}=e^{e^2} \end{align} $$

EDIT Here's something I noticed while working this out. Since all the terms are positive, we can rearrange the infinite sum as $$ \sum_{j=0}^{\infty}{\sum_{k=0}^{\infty}{\frac{k^j2^j}{k!j!}}}= \sum_{j=0}^{\infty}{\frac{2^j}{j!}\sum_{k=0}^{\infty}{\frac{k^j}{k!}}} $$ Now, the inner sum is well known. It's equal to $eB_j,$ where $B_j$ is the $j$th Bell number, so we get $$ \sum_{j=0}^{\infty}{\frac{2^j}{j!}B_j}=e^{e^2-1}. $$ This is just a special case of the exponential generating function for the Bell numbers, as described in the Wikipedia article, but substituting $x$ for $2$ in the above computation gives a third proof of the formula, that is perhaps simpler than either of the two proofs given in the article.

Answer 2

I think Weierstrass M is a good way to go for (b). For $j=1,2,\dots,$ we want to find $M_j$ such that $|f_j|\le M_j$ on $A,$ and such that $\sum_j M_j<\infty.$ WM then implies the series defining $f$ converges uniformly on $A.$ Since each $f_j$ is continuous on $A$ from (a), $f$ is continuous on $A,$ and in particular $f$ is continuous at $0$ as desired.

To find the $M_j,$ note that

$$|f_j(x)| = f_j(x)\le f_j(0)\,\, \text { for }x\in A.$$

So we take $M_j = f_j(0).$ Then

$$\sum M_j = \sum f_j(0) = \sum_{j=1}^{\infty}\sum_{k=0}^{\infty}\frac{k^j2^j}{k!j!}= \sum_{k=0}^{\infty}\sum_{j=1}^{\infty}\frac{k^j2^j}{k!j!}$$ $$ = \sum_{k=0}^{\infty}\frac{1}{k!}\sum_{j=1}^{\infty}\frac{(2k)^j}{j!} \le \sum_{k=0}^{\infty}\frac{e^{2k}}{k!} = e^{e^2} < \infty.$$

Note that switching the order of summation is justified because all terms are positive.