A sum of exponentials

Question

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 $+ \infty$?

Answer 1

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|<b $ we get $$-b<a<b $$ hence $$-1<-\frac{a+b}{2b}<0 $$ so we have $$S\left(a,b\right)=-\sum_{n=-\infty}^{\left\lfloor -\frac{a+b}{2b}\right\rfloor }e^{a+b+2nb}+\sum_{n=\left\lfloor -\frac{a+b}{2b}\right\rfloor +1}^{\infty}e^{-a-b-2nb} $$ $$=-\sum_{n=-\infty}^{-1}e^{a+b+2nb}+\sum_{n=0}^{\infty}e^{-a-b-2nb} $$ now from the well knonw identity $$\sum_{k=m}^{n}x^{k}=\frac{x^{n+1}-x^{m}}{x-1} $$ we get $$S\left(a,b\right)=-\frac{e^{a+b}}{e^{2b}-1}+\frac{e^{b-a}}{e^{2b}-1}=\color{red}{-\frac{\sinh\left(a\right)}{\sinh\left(b\right)}}$$ as wanted.