How to prove that this is a zero of the theta function?

We have the definition $$\vartheta(\tau, z) = \sum_{n=-\infty}^\infty e^{\pi i \cdot(n^2 \tau + 2 n z)}$$ and I want to show $$\vartheta(\tau, \tfrac{\tau + 1}{2}) = 0$$.

Substituting it in I get $$\begin{eqnarray} \vartheta(\tau, \tfrac{\tau+1}{2}) &=& \sum_{n=-\infty}^\infty e^{\pi i \cdot(n^2 \tau + n \tau + n)} \\ &=& \sum_{n=-\infty}^\infty e^{\pi i \cdot(n^2 \tau + n \tau)} \\ &=& \sum_{n=-\infty}^\infty e^{\pi i \cdot(n^2 + n) \cdot \tau} \\ \end{eqnarray}$$

and a hint was given that the summand is invariant under the change $$n \mapsto -(n+1)$$, indeed $$(n+1)^2 - n - 1 = n^2 + 2n + 1 - n - 1 = n^2 + n$$ so from this I understand that term $$n=0$$ equals term $$n=-1$$, term $$n=1$$ equals term $$n=2$$ and we can change the sum to:

$$\vartheta(\tau, \tfrac{\tau+1}{2}) = 2 \sum_{n=0}^\infty e^{\pi i \cdot(n^2 + n) \cdot \tau}$$

but I don't see how this is zero.

I get $$\vartheta(\tau, \tfrac{\tau+1}{2}) = \sum_{n=-\infty}^\infty e^{\pi i(n^2 \tau + n \tau + n)} = \sum_{n=-\infty}^\infty(-1)^n e^{\pi i(n^2 +n)\tau}$$ and in this sum the $$n$$ and $$-1-n$$ terms cancel each other out.
• Thank you so much! because it's $\pi i$ not $2 \pi i$!