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I want to get a general formula for $$\sum_{i = 3}^N \frac{q^{4i}}{(1-q^{2i})(1- q^{2i-4})}$$ where $q\in \mathbb{C}$ is not a root of unity (i.e. $q^n \neq 1$ for all $n\neq 0$) and $N\geq 3$. (By "formula" I mean something like $\sum_0^N i = (1+N)N/2$.)

Does anyone know if this is possible? Thanks!

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    I would try first to decompose the summand into partial fractions. – 2017-01-07
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    @Steve Assuming that the summation incxes should read i. Then the term with i = 2 makes the sum divergent for any q !=0. Maybe you made some typing error? – 2017-01-07
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    @JohnHughes: I assume $q^n \neq 1$ for any $n$. – 2017-01-07
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    @Dr.WolfgangHintze: sorry for the typo. The index should be $i$. – 2017-01-07
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    @Steve: you wrote " I assume q^n ≠1 for any n". But you can't assume this if you don't exclude n = 0. – 2017-01-07
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    @Dr.WolfgangHintze: you are correct. $q^n \neq 1$ for all $n$ but $0$. – 2017-01-07
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    @MatemáticosChibchas: I tried $\frac{q^{4i}}{(1-q^{2i})(1- q^{2i-4})} = (\frac{1}{1 - q^{2i}} - \frac{1}{1 - q^{2i - 4}})\frac{q^{2i + 4}}{q^2 - 1}$. But it seems to me this wouldn't help. – 2017-01-07
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    You still haven't addressed the $i = 2$ problem raised by Dr Hintze: when $i = 2$, the second factor of the denominator is $1 - q^{2\cdot 2 - 4} = 1 - q^0 = 1-1 = 0$. Perhaps the "general formula" is "it's undefined". :( – 2017-01-07
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    @JohnHughes: Just updated. Sorry for the overlook. I actually wanted $i$ to start from 3. – 2017-01-07
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    General MSE hint: If you can't take the time to carefully write the question you actually want answered, you lose my interest in spending *my* time answering it. It's taken more than an hour to tease out of you the thing you wanted, time that could have been spent trying to solve it. Sigh. – 2017-01-07
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    @JohnHughes: My apologizes. I should have been more careful. – 2017-01-07
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    You'll get it right next time. Best of luck. – 2017-01-07
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    @Steve: If you are studying quantum groups, then you will encounter plenty of such sums that do not have closed forms. – 2017-01-08
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    @AlexM.:Thanks! – 2017-01-08

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