I have come across the following simplification $$ g(x)=\sum_{k\in\mathbb{Z}} \frac{e^{-xk^2 }}{2k+1}=\sum_{k\in\mathbb{Z}}\frac{e^{-xk^2 }}{1-4k^2} $$ To show this, we can compute its difference $$ \sum_{k\in\mathbb{Z}} \frac{e^{-xk^2 }}{2k+1}-\sum_{k\in\mathbb{Z}}\frac{e^{-xk^2 }}{1-4k^2}=\sum_{k\in\mathbb{Z}} \frac{2ke^{-xk^2}}{4k^2-1} =0 $$ since opposite signed terms cancel out.
However, my question is, provided we do not know the right-hand side a priori, how would one go about rewriting the expression on the left to get the expression on the right? The particular motivation behind this rewriting is that, in that form, it's relatively easy to show that $g$ satisfies $$ g(x)+4g'(x)=\sum_{k\in\mathbb{Z}}e^{-xk^2 }. $$ Any ideas?
It is based on the fact that summing odd sequences yield zero, and summing even sequences can be reduced to summing over non-negative terms. So you have two ways. First one, you notice that the numerator is even, and you realize that the denominator can be transformed into an even one by multiplying both numerator and denominator by $2k-1$ and using $(a+b)(a-b)=a^2-b^2$. Then $$\sum_{k\in\mathbb Z}\frac{e^{-k^2x}}{2k+1}=\sum_{k\in\mathbb Z}\frac{(2k-1)e^{-k^2x}}{(2k+1)(2k-1)}\\=\sum_{k\in\mathbb Z}\frac{2ke^{-k^2x}}{4k^2-1}-\sum_{k\in\mathbb Z}\frac{e^{-k^2x}}{4k^2-1}$$ The first term in the sum is zero (odd), then you can incorporate the negative sign into the sum by multiplying it into the denominator.
A similar result you obtain if you write $$f(k)=\frac12\left(f(k)+f(-k)\right)+\frac12\left(f(k)-f(-k)\right)$$ The first term is even, the second is odd.