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I have the following series where $h$ and $f$ are some functions. $$1+\frac{2}{5}fh^2+\frac{4}{21}f^2h^4+...,$$ which I figured can be written as $$3\sum_{k=0} \frac{(2fh^2)^k}{4^{k+1}-1}.$$ I need help with this series. Does is it look like something specific ? Maybe it is some Taylor expansion or known series ?

user1364012
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1 Answers1

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It's not a solution, it's a hint (and too long for a clear comment).

Assume the convergence of the series, it's partly different here.

With $\displaystyle f(x):=\prod\limits_{k=0}^\infty\left(1+\frac{x}{4^k}\right)$ which has the functional equation $f(x)=(1+x)f(x/4)$ we get

$$-\frac{d}{dx}\ln f(-x)-\frac{1}{1-x}=\sum\limits_{v=0}^\infty \frac{1}{4^{v+1}-x}=\sum\limits_{v=0}^\infty \frac{1}{4^{v+1}}\sum\limits_{k=0}^\infty \frac{x^k}{4^{k(v+1)}}=$$

$$\hspace{4cm}=\sum\limits_{k=0}^\infty \frac{x^k}{4^{k+1}}\sum\limits_{k=0}^\infty \frac{1}{4^{(k+1)v}}=\sum\limits_{k=0}^\infty \frac{x^k}{4^{k+1}-1}$$

If you have a closed form for the above defined product for $f(x)$ then you will get a solution for your series.

user90369
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  • thank you for your answer! Unfortunately $x$ in my case is rather complicated and it will be not feaseble to deal with the product . – user1364012 Dec 20 '16 at 13:34
  • Sorry, that I cannot help you more, but it's hard to understand, what you like to do with $fh^2$ . There are no informations. As it's written in your post you have to calculate first $fh^2$ and then put the result into $\sum\limits_{k=0}^\infty \frac{x^k}{4^{k+1}-1}$ or equivalent into $\sum\limits_{k=0}^\infty \frac{1}{4^{k+1}-x}$. – user90369 Dec 20 '16 at 14:33
  • yes thats what it is. I probably don't understand something. If, for example, $x=e^{-t^2}$ how do I proceed ? – user1364012 Dec 20 '16 at 14:41
  • Then you have e.g. the sum $\displaystyle \sum\limits_{k=0}^\infty \frac{1}{e^{kt^2}(4^{k+1}-1)}$. Good convergence. – user90369 Dec 20 '16 at 14:44