# Summing $\frac{1}{a}-\frac{1}{a^4}+\frac{1}{a^9}-\cdots$

This question comes from temperature at sphere center. I think it's a good idea to extract the essence and post a pure mathematical question to attract more thoughts. It is a physical problem and interested readers can go to the original post to find details.

Anyway, after simplification the wanted value is $2 f(x)$

$$f(x)= - \sum_{n=1}^{\infty}(-1)^n e^{-x n^2}$$ Letting $x = \pi^2 D t /a^2$ gives the answer to the original question. If we further let $e^{x} = a$, we have a summation problem:

$$S=-\sum_{n=1}^{\infty}(-1)^n a^{-n^2} = \frac{1}{a}-\frac{1}{a^4}+\frac{1}{a^9}-\cdots$$

$a > 1$ so $S$ converges, but I don't now how to sum it. Any suggestions?

• @i8Σπ_821 Could you be more specific? That's not a geometric series. Mar 30, 2017 at 2:18
• Ah yes it's not. I will delete my comment. Apologies. Mar 30, 2017 at 2:24
• @i8Σπ_821 Hi, no problem :-) Mar 30, 2017 at 2:28
• Taozi: I've never heard of that function that Misha mentioned, but it's interesting :-) Mar 30, 2017 at 2:32
• @i8Σπ_821 I just posted an answer to the original physical problem. The link to is in the first sentence of my question. I plotted a figure there. Mar 30, 2017 at 2:34

## 1 Answer

We can write this sum in terms of the Jacobi theta function $\vartheta(z;\tau)$; in particular, I believe that $$\vartheta(\tfrac12;\tfrac{ix}{\pi}) = 1 - 2f(x).$$ That's not an answer: it's just saying "we don't know how to find this sum in terms of functions we know about, so we gave this sum a name".

But maybe looking at the Wikipedia article I linked to will give you some idea of what can be done with the function to find the things about it you're interested in.

• Thanks a lot, that's a beautiful function that I've never heard of. Mar 30, 2017 at 2:17