# sum of series of the form $\sum\limits_{n=-\infty}^{\infty}-\frac{(1+2n)}{2}\exp(-\frac{(1+2n)^2}{4})$.

Consider the series

$\sum\limits_{n=-\infty}^{\infty}-\frac{(1+2n)}{2}\exp(-\frac{(1+2n)^2}{4})$.

By using Mathematica I found the sum of the series is zero.

Question: I can Show that the series is convergent but I couldn't find its sum. Could anyone please help to find the sum of the series as $0$ ? Please help me, I have really no idea to find its sum.

• The zeroth term is the negative of the $n=-1$ term. There are a lot of cancellation. – user99914 Mar 25 '15 at 7:46
Taking into account John's comment, consider $$S_p=\sum\limits_{n=-p}^{p}-\frac{(1+2n)}{2}~\exp(-\frac{(1+2n)^2}{4})$$ and write a few terms; as John says, there are a lot of cancelations and you are let with $$S_p=-\frac{(1+2p)}{2}~\exp(-\frac{(1+2p)^2}{4})$$ Now consider $x=\frac{(1+2p)}{2}$ which makes $S_p$ looking like $-x~e^{-x^2}$ and make $x$ infinitely large.