Since:
![\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \sum_{v=1}^{\infty} \frac{(-1)^{v+1}}{v} \left[-\frac{1}{n^{2}m^{2}\pi^{2}}\right]^{v}](http://s0.wp.com/latex.php?latex=%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D+%5Csum_%7Bm%3D1%7D%5E%7B%5Cinfty%7D+%5Csum_%7Bv%3D1%7D%5E%7B%5Cinfty%7D+%5Cfrac%7B%28-1%29%5E%7Bv%2B1%7D%7D%7Bv%7D+%5Cleft%5B-%5Cfrac%7B1%7D%7Bn%5E%7B2%7Dm%5E%7B2%7D%5Cpi%5E%7B2%7D%7D%5Cright%5D%5E%7Bv%7D&bg=1b1b1b&fg=dddddd&s=0 "\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \sum_{v=1}^{\infty} \frac{(-1)^{v+1}}{v} \left[-\frac{1}{n^{2}m^{2}\pi^{2}}\right]^{v}")
![\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \sum_{v=1}^{\infty} \frac{(-1)^{2v+1}}{v} \left[\frac{1}{n^{2}m^{2}\pi^{2}}\right]^{v}](http://s0.wp.com/latex.php?latex=%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D+%5Csum_%7Bm%3D1%7D%5E%7B%5Cinfty%7D+%5Csum_%7Bv%3D1%7D%5E%7B%5Cinfty%7D+%5Cfrac%7B%28-1%29%5E%7B2v%2B1%7D%7D%7Bv%7D+%5Cleft%5B%5Cfrac%7B1%7D%7Bn%5E%7B2%7Dm%5E%7B2%7D%5Cpi%5E%7B2%7D%7D%5Cright%5D%5E%7Bv%7D&bg=1b1b1b&fg=dddddd&s=0 "\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \sum_{v=1}^{\infty} \frac{(-1)^{2v+1}}{v} \left[\frac{1}{n^{2}m^{2}\pi^{2}}\right]^{v}")
(which can easily be evaluated in wolfram alpha, and converges much faster than the original double product)
Therefore:
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Since:
(which can easily be evaluated in wolfram alpha, and converges much faster than the original double product)
Therefore:
