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i tried a few thing but none them worked out so I'm totally clueless what to do to solve this $$\sum_{n=0}^{\infty}\frac{x^n}{A(x^{2n}-1)+1}$$

Zev Chonoles
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mh96
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    What is $A$? ${}$ – Zev Chonoles Jul 27 '13 at 18:51
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    ...and what is $,x,$ ...? – DonAntonio Jul 27 '13 at 18:52
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    their both constant numbers so i think the answer should be in the form of $$F(x,A)$$ – mh96 Jul 27 '13 at 18:58
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    The ratio of terms is:

    $$ a_{n+1}/a_n = x\frac {Ax^{2n} -A +1} {Ax^{2n+2}-A +1}.$$

    If $|x|\leq1$, this goes to $x$. If $|x|>1$, this goes to $1/x$. Therefore it converges by the ratio test for all $x$, $|x|\neq 1$. We test at $x = 1, -1$. By inspection, it diverges for these. So it converges exactly on $${x : |x|\neq 1 }.$$

     – Eric Auld Jul 27 '13 at 19:07
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    This question is a valid and interesting one. The partial sums of this series can be expressed in closed form in terms of the $q$-polygamma function (http://mathworld.wolfram.com/q-PolygammaFunction.html). Then probably the sum can be computed taking limits. I am not familiar with these functions but some around here are and will give a solution from which we will all learn something interesting. There is no need for patronizing the OP with stupid questions like what is $A$ and $x$. – OR. Jul 27 '13 at 19:14
  • @RGB, I don't think it was obvious what $A$ was. (I thought it was a function at first, until the OP corrected us.) – George V. Williams Jul 27 '13 at 19:22
  • oh, that is right. My apologies to Chonoles. – OR. Jul 27 '13 at 19:26

1 Answers1

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This is only a bit of unjustified information

It seems to be (by experimenting with different values of $A$) that the partial sums ($\sum_{n=1}^{m}$) of the series are

$-\frac{\psi_{x}\left(m-\frac{i\pi-\frac{1}{2}\ln(\frac{A}{A-1})}{\ln(x)}+1\right)}{2\sqrt{A(A-1)}\ln(x)}+\frac{\psi_{x}\left(m-\frac{\frac{1}{2}\ln(\frac{A}{A-1})}{\ln(x)}+1\right)}{2\sqrt{A(A-1)}\ln(x)}+\frac{\psi_{x}\left(1-\frac{i\pi-\frac{1}{2}\ln(\frac{A}{A-1})}{\ln(x)}+1\right)}{2\sqrt{A(A-1)}\ln(x)}-\frac{\psi_{x}\left(1-\frac{\frac{1}{2}\ln(\frac{A}{A-1})}{\ln(x)}+1\right)}{2\sqrt{A(A-1)}\ln(x)}$

where $\psi_{q}(z)$ is the $q$-polygamma function. For those familiar with playing with these functions this may help/give an idea, perhaps.

I experimented with the summation starting from $n=1$ by mistake. The difference is just the first term anyway.

OR.
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