# Summation of exponential terms

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1. Mar 23, 2015

### A_s_a_d

I found the following identity in a paper:
$\sum_{l=1}^{\infty}exp(-\pi\alpha l^2)=(\frac{1}{2\sqrt{\alpha}}-\frac{1}{2})+\frac{1}{\sqrt{\alpha}}\sum_{l=1}^{\infty}exp(\frac{-\pi l^2}{\alpha})$
Someone please let me give some hints on how to prove this.

2. Mar 23, 2015

### jfizzix

It looks like it's the same sum on both sides, but scaled.

I would see what happens to the exponential sum when you change $\alpha$ to $\alpha\beta$. Can you make the result identical with the right scaling factor in front (twice as wide, but half as tall)?

3. Mar 27, 2015

### A_s_a_d

Yes, it's the same sum but scaled. One thing I have found that if you approximate the summation as integral, it can be proved easily as both are usual Gaussian Integral. But I was worrying about the factor that involves to transform the summation into integral. Any thoughts?

4. Mar 27, 2015

### jfizzix

So, you can write it alternatively as:
$\frac{1}{2} +\sum_{l=1}^{\infty}exp(-\pi\alpha l^2)=\frac{1}{\sqrt{\alpha}}\big(\frac{1}{2}+\sum_{l=1}^{\infty}exp(\frac{-\pi l^2}{\alpha})\big)$
or better yet
$\sum_{l=-\infty}^{\infty}exp(-\pi\alpha l^2)=\frac{1}{\sqrt{\alpha}}\sum_{l=-\infty}^{\infty}exp(\frac{-\pi l^2}{\alpha})$
this sum is exactly the integral of a discrete gaussian. With the normalization constant in front, they should have the same area..

At least, that is how I think it should work out.

Does $\alpha$ have to be a positive integer?