Is this formula for $\sum_{n} (n^{2}+z^{2})^{-s}$ correct?

I would like to know if this formula is true:

$$\sum_{n=1}^{\infty}\frac{1}{(z^{2}+n^{2})^s}=\frac{1}{\Gamma(s)} \sum_{n=0}^{\infty}\Gamma(s+n)\zeta(2s+2n)\frac{ (-z^2)^n}{n!}.$$

I have used the fact that

$$\Gamma(s)\sum_{n=1}^{\infty}\frac{1}{(z^{2}+n^{2})^s}=\int_{0}^{\infty}t^{s-1}\exp(-tz^{2})\sum_{n=1}^{\infty}\exp(-tn^{2})~dt$$

and that the Mellin transform of $\displaystyle \sum_{n=1}^{\infty}\exp(-tn^{2})$ is just $\Gamma(s) \zeta (2s)$.

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As long as the interchange of limits is valid (I'm horrible at checking that), it should be true wherever convergence isn't an issue. I don't see any mistakes. –  anon Jun 26 '12 at 11:01
I don't see how it can be. When $z=0$ the left side gives $\zeta(2s)$ and the right gives zero. –  Zander Jun 26 '12 at 16:37
@anon The original question had $z^2$ in the right side summand but your edit has $z^{2n}$. Was that intentional? –  Zander Jun 26 '12 at 16:44
@Zander: Yes, Jose had a typo. (It comes from expanding $\exp(-tz^2)$ in the second line.) Also, $0^0=1$ in the contexts of polynomials and power series expansions, so in fact you get $\zeta(2s)=\zeta(2s)$, as expected. –  anon Jun 26 '12 at 16:46