I find that every so often I end up with a double sum that could be changed into a different form and make some expression simpler. I have a situation I am in right now that should be able to benefit but I am not sure how to rigorously go about making an appropriate change of variables. In particular, I have the following sum:
$$\frac{1}{2}\sum_{\substack{n = 1 \\n \text{ odd}}}^{\infty} \sum_{k=0}^n \frac{x^k}{k!} \frac{x^{n-k}}{(n-k)!}$$
and I want to show that it is equivalent to
$$\left(\sum_{\substack{i = 1 \\i \text{ odd}}}^\infty \frac{x^i}{i!}\right) \left(\sum_{\substack{j = 0 \\j \text{ even}}}^\infty \frac{x^j}{j!}\right)$$
How can I do a change of index variables to go from the first equation to the latter and vice versa? My thoughts to go from the second equation to the first is to observe that $ 1 \leq i \leq \infty$ and $0 \leq j \leq \infty$ which implies that $1 \leq i+j \leq \infty$, so choosing $i+j=n$ gives us that $1 \leq n \leq \infty$ and that $n$ is odd since $i+j$ must always be odd. Then it seems I can maybe use the fact that $1 \leq i \leq n \leq \infty$ and $0 \leq j \leq n \leq \infty$ to say that $k$ should be the union of $i$ and $j$ up to $n$, implying that $0 \leq k \leq n$.
The above does not feel rigorous enough though, so can anyone suggest a better way to, in general, go between these different index variables in situations like this? Should I be trying to find some bijection between the index sets and if so, is there a clear way to do this?
[Edit 1]
After revisiting this problem, I used steps similar to above except found I made a logical error. First, consider the following simplification
\begin{align} \left(\sum_{\substack{i = 1 \\i \text{ odd}}}^\infty \frac{x^i}{i!}\right) \left(\sum_{\substack{j = 0 \\j \text{ even}}}^\infty \frac{x^j}{j!}\right) &= \sum_{\substack{i = 1 \\i \text{ odd}}}^\infty \sum_{\substack{j = 0 \\j \text{ even}}}^\infty \frac{x^{i+j}}{i! j!} \\ &= \sum_{\substack{i = 1 \\i \text{ odd}}}^\infty \sum_{\substack{j = 0 \\j \text{ even}}}^\infty \binom{i+j}{i}\frac{x^{i+j}}{(i+j)!} \\ \end{align}
With this simplification in hand, do a similar analysis to above where we use the fact that $1 \leq i \leq \infty$ and $0 \leq j \leq \infty$ to find that $1 \leq i+j \leq \infty$ and then choosing a new variable $n = i+j$ which is known to be odd since $i$ is odd and $j$ is even. We then know that $1 \leq i \leq n$ with $i$ remaining odd, which gives us that
\begin{align} \sum_{\substack{i = 1 \\i \text{ odd}}}^\infty \sum_{\substack{j = 0 \\j \text{ even}}}^\infty \binom{i+j}{i}\frac{x^{i+j}}{(i+j)!} &= \sum_{\substack{n = 1 \\n \text{ odd}}}^\infty \sum_{\substack{i = 1 \\i \text{ odd}}}^n\binom{n}{i}\frac{x^{n}}{n!} \\ &= \sum_{\substack{n = 1 \\n \text{ odd}}}^\infty \frac{x^{n}}{n!} \sum_{\substack{i = 1 \\i \text{ odd}}}^n\binom{n}{i} \end{align}
The last observation comes from the inner sum $\sum_{\substack{i = 1 \\i \text{ odd}}}^n\binom{n}{i}$. Since $n$ is odd, we know there is an even number of integers from $0$ to $n$, half of which are odd and half which are even. If we consider the sum $\sum_{\substack{i = 0 \\i \text{ even}}}^n\binom{n}{i}$, we can recognize that
\begin{align} \sum_{\substack{i = 0 \\i \text{ even}}}^n\binom{n}{i} &= \sum_{\substack{i = 0 \\i \text{ even}}}^n\binom{n}{n-i} \\ &= \sum_{\substack{j = 1 \\j \text{ odd}}}^n\binom{n}{j} \end{align}
where we noticed that $n-i = j$ is an odd number for each $i$ in that sum. Thus, this sum is equivalent to the inner sum we already have and we know that
$$\sum_{\substack{i = 0 \\i \text{ even}}}^n\binom{n}{i} + \sum_{\substack{i = 1 \\i \text{ odd}}}^n\binom{n}{i} = \sum_{i=0}^n\binom{n}{i} $$
implying that
$$\sum_{\substack{i = 1 \\i \text{ odd}}}^n\binom{n}{i} = \frac{1}{2}\sum_{i=0}^n\binom{n}{i}$$
This then gives us that our sum of interest is equal to
\begin{align} \sum_{\substack{i = 1 \\i \text{ odd}}}^\infty \sum_{\substack{j = 0 \\j \text{ even}}}^\infty \binom{i+j}{i}\frac{x^{i+j}}{(i+j)!} &= \frac{1}{2}\sum_{\substack{n = 1 \\n \text{ odd}}}^\infty \frac{x^{n}}{n!} \sum_{i=0}^n\binom{n}{i} \\ &= \frac{1}{2}\sum_{\substack{n = 1 \\n \text{ odd}}}^\infty \sum_{i=0}^n \frac{x^i}{i!} \frac{x^{n-i}}{(n-i)!} \end{align}
which shows the desired equality. Thus, my original thought that transforming the summation indicies was the only thing necessary to arrive at equality was not correct, one also needed to make use of the unimodality of binomial coefficients.
In the first expression, the indexes $nk$ follow the pattern
$$10,11,30,31,32,33,50,51,52,53,54,55,\cdots$$
and the reversal with $n-k$ does not matter.
In the second expression, the terms are
$$1,3,5,\cdots\times 0,2,4,\cdots$$ hence
$$10,12,14,\cdots 30,32,34,\cdots 50,52,54,\cdots$$
For comparison, you can "fold" the pairs so that $k\le n$ and get
$$10,21,41,\cdots 30,32,43,\cdots 50,52,54,\cdots$$ See the mismatch.
To solve such problems, you can sketch the domain of the index pairs in the plane.