I'm trying to evaluate the series: $$\sum_{n=1}^\infty \frac{1}{3^{2^n}-3^{-2^n}}$$ I have tried to put it into partial fractions but it doesn't seem to telescope. Does anyone have any ideas?
According to Wolfram, the answer is $\frac{1}{8}$.
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Strategy: find an expression that can be replaced by a geometric series. Here, \begin{align*} \sum_{n=1}^\infty \frac{1}{3^{2^n}-3^{-2^n}} &= \sum_{n=1}^\infty 3^{-2^n} \frac{1}{1-3^{-2^{n+1}}} \\ &= \sum_{n=1}^\infty 3^{-2^n} \sum_{k=0}^\infty 3^{-2^{n+1}k} = \sum_{n=1}^\infty \sum_{k=0}^\infty 3^{-2^n(2k+1)}. \end{align*} All the exponents are even, and every even number $2m$ can be written uniquely as $2m=2^n(2k+1)$ where $n\ge1,k\ge0$. Therefore $$ \sum_{n=1}^\infty \sum_{k=0}^\infty 3^{-2^n(2k+1)} = \sum_{m=1}^\infty 3^{-2m} = \frac{1/9}{1-1/9} = \frac18 $$ (another geometric series).
Greg Martin
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Greg Martin's answer is very concise and elegant. Interesting to know that $2^n(2k+1)$ cycles through all even numbers.
Here's my attempt - a much less elegant one. Partial fractions comes in handy but surprisingly not for telescoping.
Hypergeometricx
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$$\sum_{n=1}^N \frac{1}{3^{2^n}- 3^{-2^n}} = \frac{1}{8} - \frac{1}{3^{2^{N+1}}-1}.$$
– Daniel Fischer Sep 26 '14 at 23:42