How to solve $\sum_{k=2}^\infty {\frac{1}{k^2-1}}$
Solution 1:
HINT
$$\dfrac1{k^2-1} = \dfrac12 \left(\dfrac1{k-1} - \dfrac1{k+1} \right)$$ Write out some terms and you can notice some nice cancelling going on.
If you want to stick to your integral test, then note that $\ln(M-1) - \ln(M+1) = \ln \left(\dfrac{M-1}{M+1}\right)$ which tends to $\ln(1) = 0$ for large $M$.
Solution 2:
Simplify that last expression:
$$\ln|M-1|-\ln|M+1|=\ln\left|\frac{M-1}{M+1}\right|=\ln\left|1-\frac2{M+1}\right|\to 0$$ as $M\to\infty$. (Actually, you can evaluate the sum exactly, since the series is two intermeshed telescoping series.)