How do you evaluate:
$\displaystyle \sum_{n=0}^{\infty}x^{n^{2}} $
Or more generally
$ \large\displaystyle \sum_{n=0}^{\infty}x^{n^{\alpha}} $
Note that: $|x| <1$
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Kunal Gupta
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Since you added the tag, you probably know that the first summation is $\frac{1}{2} (1+\vartheta _3(0,x))$. For the general case, I don't know if there is any closed form. – Claude Leibovici Nov 01 '15 at 14:32
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1There is no closed form for a general $\alpha$, but we do have this interesting property. – Lucian Nov 01 '15 at 17:43
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There is the Jacobi Theta function $$ \vartheta_3(z,q) = \sum_{n=-\infty}^\infty e^{2 i n z} q^{n^2} $$
so yours is $$ \sum_{n=0}^\infty x^{n^2} = \frac{\vartheta_3(0,x)+1}{2} $$
Of course for $a=1$ it can also be evaluated using known functions. But not for other values of $a$.
GEdgar
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