I'm wondering if there is a closed formula for the sum $a^1+a^4+a^9...$ and more generally $a^{1^n}+a^{2^n}+a^{3^n}...$ for real $a$ and $n$ such that $|a|<1$ and $n>1$.
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$\begingroup$ Depends what you mean by "closed formula" :P $\endgroup$– Rivers McForgeCommented Jul 23, 2020 at 3:58
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$\begingroup$ @RiversMcForge Without iterated sums or products, but honestly anything easier to evaluate. I feel like it's a Taylor series of something but that doesn't seem like the case since Taylor series powers grow linearly. $\endgroup$– Kevin LuCommented Jul 23, 2020 at 4:10
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$\begingroup$ Since this is a special case of the Jacobi theta function, $a + a^4 + a^9 + a^{16} + ... = \vartheta(0; \frac{\ln a}{\pi i})$, I doubt it has a "nice" closed form in terms of "elementary" functions. I can make this into an answer if you want. en.wikipedia.org/wiki/Theta_function $\endgroup$– Rivers McForgeCommented Jul 23, 2020 at 4:15
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1 Answer
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According to Jacobi Triple product formula we have
$$\prod_{m=1}^{\infty}(1-x^{2m})(1+x^{2m-1}y^2)\left(1+\frac{x^{2m-1}}{y^2}\right)=\sum_{n=-\infty}^{\infty}x^{n^2}y^{2n}$$ for $|x|<1$ and $y\neq0$. Then taking $x=a$ and $y=1$ we get $$\prod_{m=1}^{\infty}(1-a^{2m})(1+a^{2m-1}1^2)\left(1+\frac{x^{2m-1}}{1^2}\right)=\sum_{n=-\infty}^{\infty}a^{n^2}1^{2n}$$
$$\implies\prod_{m=1}^{\infty}(1-a^{2m})(1+a^{2m-1})^2=2\sum_{n=1}^{\infty}a^{n^2}+1$$
$$\implies\sum_{n=1}^{\infty}a^{n^2}=\frac{\prod_{m=1}^{\infty}(1-a^{2m})(1+a^{2m-1})^2-1}{2}$$
You can find more things here http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.68.6437&rep=rep1&type=pdf
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2$\begingroup$ @SubhrajitBhattacharya: But that's not the question as asked... right? $\endgroup$ Commented Jul 23, 2020 at 3:21
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1$\begingroup$ No. But I obtained a a formula using Jacobi Triple product and therefore posted. $\endgroup$– ShBhCommented Jul 23, 2020 at 3:22
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1$\begingroup$ Well, I can generate an unrelated equation using a different principle. does that mean I should post that???!? $\endgroup$ Commented Jul 23, 2020 at 4:55