Limit of $\sum_{k=2}^{\infty} \frac{1}{k^2-1}$

Question

My calculator suggests that the limit of this series is 0.75. $$\sum_{k=2}^{\infty} \frac{1}{k^2-1}$$ How can this be proved. I was thinking about a telescop sum but that didn't help me I appreciate any help you could provide

Answer 1

The sum can be rewritten as a telescoping one! Notice

$$\begin{align}\frac{1}{k^2-1} = \frac{1}{(k-1)(k+1)} & \stackrel{\color{blue}{[1]}}{=} \frac12\left(\frac{1}{k-1} - \frac{1}{k+1}\right) \\ & \stackrel{\color{blue}{[2]}}{=} \frac12\left[\left(\frac{1}{k-1}+\frac{1}{k}\right) - \left(\frac{1}{k} + \frac{1}{k+1}\right)\right]\\ &= \frac12\left(\frac{2k-1}{(k-1)k} - \frac{2k+1}{k(k+1)}\right) \end{align} $$ This leads to

$$\begin{align}\sum_{k=2}^\infty \frac{1}{k^2-1} &= \lim_{p\to\infty}\sum_{k=2}^p\frac{1}{k^2-1} = \frac12\lim_{p\to\infty}\sum_{k=2}^p\left(\frac{2k-1}{(k-1)k} - \frac{2k+1}{k(k+1)}\right)\\ &= \frac12\lim_{p\to\infty}\left(\frac{2(2)-1}{(2-1)2} - \frac{2p+1}{p(p+1)}\right) = \frac12\left(\frac32 - 0\right) = \frac34\end{align}$$

Notes

• $\color{blue}{[1]}$ This is a special case of the identity $\frac{1}{(x+a)(x+b)} = \frac{1}{b-a}\left(\frac{1}{x+a} - \frac{1}{x+b}\right)$.
  A lot of summands in this sort of sums can be simplified using this.

• $\color{blue}{[2]}$ This is simply filling the missing terms between differences of terms from same arithmetic progression.This trick is useful to turn something that looks like telescoping to an actual telescoping one.

Answer 2

$\sum_{k=2}^{m}\dfrac{1}{k^2-1}=$

$(1/2)\sum_{k=2}^{m}\dfrac{1}{k-1} -$

$(1/2)\sum_{k=2}^{m}\dfrac{1}{k+1}=$

$(1/2)\sum_{k=1}^{m-1}\dfrac{1}{k}-$

$(1/2)\sum_{k=3}^{m+1}\dfrac{1}{k}=$

$(1/2)[1+1/2] -$

$ (1/2)[1/m+1/(m+1)].$

$\lim_{ m \rightarrow \infty} S_m = 3/4.$