Show that this series converges and find its sum.

Question

$$\sum_{n=1}^{\infty} \frac{1}{\cosh(n)\cosh(n+1)}$$

For showing that it is convergent. I prove it by using ratio test. But for finding the value of sum. I stuck on this part.

Answer 1

\begin{align} \sum_{n=1}^\infty \frac{1}{\cosh(n) \cosh (n+1)} &= \sum_{n=1}^\infty \frac{1}{\left(\frac{e^n + e^{-n}}{2}\right)\left(\frac{e^{n+1} + e^{-(n+1)}}{2}\right)}\\ &= \sum_{n=1}^\infty \frac{4}{\left(e^n + e^{-n}\right)\left(e^{n+1} + e^{-(n+1)}\right)}\\ &= 4 e\sum_{n=1}^\infty \frac{1}{e^{-2 n}+e^{2 n+2}+e^2+1}\\ &= \frac{4 e}{1-e^2} \sum_{n=1}^\infty \left(\frac{1}{e^{2n+2}+1} - \frac{1}{e^{2 n}+1}\right) \\ &= \frac{4 e}{1-e^2} \left[\left(\frac{1}{e^{4}+1} + \frac{1}{e^{6}+1} + \dots\right) - \left(\frac{1}{e^{2}+1} + \frac{1}{e^{4}+1} \dots\right)\right] \\ &= \frac{4 e}{1-e^2} \left(-\frac{1}{e^2+1}\right) \\ &= \frac{4 e}{e^4-1} \end{align}