How to solve $\sum_{k=2}^\infty {\frac{1}{k^2-1}}$ I'm using the integral test to determine if this series converges. From what I have so far it seems that it diverges, but according to wolfram alpha it converges. Where is my mistake? 
$$\sum_{k=2}^\infty {\frac{1}{k^2-1}} \to \int_{2}^M \frac{1}{x^2-1}dx$$
$$\int_{2}^M \frac{1}{x^2-1}dx = \frac{1}{2}\Big(\ln|M-1|-\ln|M+1|+\ln|3|\Big)$$
$$ \lim_{M\to \infty} \frac{1}{2}\Big(\ln|M-1|-\ln|M+1|+\ln|3|\Big)$$
 A: 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 $M$.
A: 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.)
A: The integral test, used directly, works, if one can evaluate the limit. However, many functions are either very difficult to integrate, or have an integral that cannot be expressed in terms of elementary functions.   
So the first step probably should have been a simplification. Note that for $k\ge 2$, we have $k^2-1\gt \frac{k^2}{2}$, and therefore 
$$0\lt \dfrac{1}{k^2-1}\lt \dfrac{2}{k^2}.$$
Now you probably know that $\displaystyle\sum_{k=2}^\infty \dfrac{1}{k^2}$ converges, so by comparison our series does.
If one doesn't know that (unlikely at this stage), the integral test can be used on $\displaystyle \sum_2^\infty \dfrac{2}{k^2}$.
