it vanishes! (more playing around with zeta and theta functions)

So I’m still thinking about zeta and theta functions, and came up with this observation:
\sum_{n=2}^{\infty} \frac{\theta_3(0,1/n^2)}{n^2}

\sum_{n=2}^{\infty} \frac{\theta_3(0,1/n^2)}{n^2}= \sum_{n=2}^{\infty} \frac{1}{n^2} \sum_{m=-\infty}^{\infty} \left(\frac{1}{n^2}\right)^{m^2}

\sum_{n=2}^{\infty} \frac{1}{n^2} \sum_{m=-\infty}^{\infty} \left(\frac{1}{n^2}\right)^{m^2} = \sum_{m=-\infty}^{\infty}  \sum_{n=2}^{\infty} \frac{1}{n^2} \left(\frac{1}{n^2}\right)^{m^2} = \sum_{m=-\infty}^{\infty}  \sum_{n=2}^{\infty} \frac{1}{n^{2+2m^2}}

\sum_{m=-\infty}^{\infty}  \sum_{n=2}^{\infty} \frac{1}{n^{2+2m^2}} = \sum_{m=-\infty}^{\infty} \zeta(2+2m^{2})-1

Just a little test with mpmath:

>>> nsum(lambda n: jtheta(3,0,1/(n*n))/(n*n), [2,inf])
>>> nsum(lambda m: zeta(2*m*m+2)-1, [-inf,inf])

The same trick can be applied to
\sum_{n=2}^{\infty} \frac{\theta_3(0,1/n)}{e^{n}} = \sum_{n=2}^{\infty} \frac{1}{e^{n}} \sum_{m=-\infty}^{\infty} \left(\frac{1}{n}\right)^{m^2}

\sum_{n=2}^{\infty} \frac{1}{e^{n}} \sum_{m=-\infty}^{\infty} \left(\frac{1}{n}\right)^{m^2} = \sum_{m=-\infty}^{\infty}  \sum_{n=2}^{\infty} \frac{1}{e^{n}} \left(\frac{1}{n}\right)^{m^2} = \sum_{m=-\infty}^{\infty}  \sum_{n=2}^{\infty} \frac{(1/e)^{n}}{n^{m^2}}

\sum_{m=-\infty}^{\infty}  \sum_{n=2}^{\infty} \frac{(1/e)^{n}}{n^{m^2}}=\sum_{m=-\infty}^{\infty} [\mathrm{Li}_{m^{2}}(1/e)-1/e ]

>>> nsum(lambda n: jtheta(3,0,1/n)/exp(n), [2,inf])
>>> nsum(lambda m: polylog(m*m,1/e)-1/e, [-inf,inf])

I think that when we see the Riemann zeta function most of the time we’re looking at the surface and not at the plumbing, and the plumbing of the Riemann zeta function is rife with theta functions — this entry, my previous entry about the alternate representation of the product, and the regularized Mellin transform integral definition for the Riemann zeta function.


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