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Is there any closed form for the finite sum $$\sum_{k=1}^n\dfrac{1}{r^{k^2}}$$ or infinite sum ( when $|r|<1$) $$\sum_{k=1}^\infty\dfrac{1}{r^{k^2}} ?$$ While solving this problem, I found this type of finite series. But I have no idea about attempt to this problem. Thank you.

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Bumblebee
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    The sum can be expressed as a theta function, and that is probably the best you can do. – Winther Jan 24 '15 at 13:35
  • You got a good suggestion by the previous comment. – Mhenni Benghorbal Jan 24 '15 at 13:56
  • @Winther: Thank you. You show me a new direction. But I am not familiar with theta function. Please can you explain little bit more? – Bumblebee Jan 24 '15 at 14:36
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    @Nilan. Same for me ... but nice pictures ! – Claude Leibovici Jan 24 '15 at 14:44
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    Theta functions are horrible functions with beautiful properties. Writing down the solution in terms of theta functions ($=[\vartheta(0;\log(1/r)/i\pi)-1]/2$) gives you a closed form solution that is only that. I don't think it will be very useful for calculating the sum, though it does give you access to use known properties of theta functions if you are going to continue doing stuff with the sum. – Winther Jan 24 '15 at 14:45
  • A more useful answer to your question is that your sum has no useful (in terms of evaluating it) closed form solution. – Winther Jan 24 '15 at 14:54
  • @Nilan: I have posted a partial answer to your question. – DER Jan 29 '15 at 16:48

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We have

$$\sum_{k=1}^2 1/r^{k^2}= (r^3+1)/r^4$$ $$\sum_{k=1}^3 1/r^{k^2}= (r^8+r^5+1)/r^9$$ $$\sum_{k=1}^4 1/r^{k^2}= (r^{15}+r^{12}+r^7+1)/r^{16}$$ $$\sum_{k=1}^5 1/r^{k^2}= (r^{24}+r^{21}+r^{16}+r^9+1)/r^{25}$$

So, for a fixed $n$ we have

$$\sum_{k=1}^n 1/r^{k^2}= w_{n-1}(r)/r^{n^2}$$

where $w_{n-1}(r)$ is a polynomial of degree $n^2-1$ in $r$. But obtaining the formula when $n$ goes to infinity is impossible as hinted in the above comments.

Ron Kaminsky
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DER
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