Sum $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{ch(n)}{3^n}$

Question

Find a sum of $$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{ch(n)}{3^n}$$

Could you give some some hint or some way to start this? I have tried representing ch(n) through its definition with e, but I can't find any pattern that would tell me the n-th therm of the partial sum.

Answer 1

Assuming $\text{ch}\, x = \cosh x = \frac{e^x + e^{-x}}{2}$, write $$\sum_{n=1}^\infty (-1)^{n+1} \frac{\text{ch}(n)}{3^n}= -\frac{1}{2}\sum_{n=1}^\infty \left(-\frac{e}{3}\right)^n + \left(-\frac{1}{3e}\right)^n,$$ which should be easy to evaluate.

Answer 2

Hint

$$(-1)^{n+1}\frac{\cosh(n)}{3^n}=-\frac12\left(\left(\frac{-e}{3}\right)^n+\left(\frac{-1}{3e}\right)^n\right)$$ and $$\sum_{n=1}^\infty x^n=\frac{x}{1-x},\quad|x|<1$$