How would you prove that $$\sum_{n \ \text{odd}} \text{sgn}(a + nb ) e^{- | a + n b |} = - \frac{\sinh(a)}{\sinh(b)}$$
where $|a| < b$ and the sum runs over all odd integers between $-\infty$ and $+ \infty$?
A sum of exponentials
3
$\begingroup$
sequences-and-series
analysis
summation
hypergeometric-function
hyperbolic-functions
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0We can write: $$\sum_{\text{n}\space\text{odd}}\frac{\text{sgn}\left(\text{a}+\text{n}\text{b}\right)}{\exp\left(\left|\text{a}+\text{n}\text{b}\right|\right)}=$$ $$\sum_{\text{n}=-\infty}^0\frac{\text{sgn}\left(\text{a}+\left(2\text{n}+1\right)\text{b}\right)}{\exp\left(\left|\text{a}+\left(2\text{n}+1\right)\text{b}\right|\right)}+\sum_{\text{n}=1}^\infty\frac{\text{sgn}\left(\text{a}+\left(2\text{n}+1\right)\text{b}\right)}{\exp\left(\left|\text{a}+\left(2\text{n}+1\right)\text{b}\right|\right)}$$ – 2017-01-07
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0Yes, that's true. – 2017-01-07
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0For the right sum (that starts at $\text{n}=1$ and go to $\infty$), you can say (using the ratio test): $$\exp\left[-2\sqrt{\Re^2\left(\text{b}\right)+\Im^2\left(\text{b}\right)}\right]<1$$ – 2017-01-07
1 Answers
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Note that we may write the series as $$S\left(a,b\right)=\sum_{n\in\mathbb{Z}}\textrm{sgn}\left(a+\left(2n+1\right)b\right)e^{-\left|a+\left(2n+1\right)b\right|} $$ now note that $$a+\left(2n+1\right)b>0\Leftrightarrow n>-\frac{a+b}{2b}. $$ Since $\left|a\right|