Example

1. Find the sum $$\sum_{k=2}^n \dfrac{2}{k^2-1}.$$
2. Does the infinite series $$\sum_{k=2}^\infty \dfrac{2}{k^2-1}$$ have a limiting sum? If so, what is its value?

Solution

1. We can factor $k^2-1$ and split the summand into $$\dfrac{2}{k^2-1} = \dfrac{1}{k-1} - \dfrac{1}{k+1}.$$

   Thus,

   $$\begin{align*} \sum_{k=2}^n \dfrac{2}{k^2-1} &= \sum_{k=2}^n \Bigl( \dfrac{1}{k-1} - \dfrac{1}{k+1} \Bigr) \\ &= \sum_{k=2}^n \dfrac{1}{k-1} - \sum_{k=2}^n \dfrac{1}{k+1}. \end{align*}$$ If we write out the terms of these two sums, we have $$\Bigl(1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots + \dfrac{1}{n-2} + \dfrac{1}{n-1} \Bigr) - \Bigl(\dfrac{1}{3} + \dfrac{1}{4} + \dots + \dfrac{1}{n-2} + \dfrac{1}{n-1} + \dfrac{1}{n} + \dfrac{1}{n+1}\Bigr).$$ Most of the terms cancel out (telescope), giving $$\begin{align*} \sum_{k=2}^n \dfrac{2}{k^2-1} &= 1 + \dfrac{1}{2} - \dfrac{1}{n} - \dfrac{1}{n+1} \\ &= \dfrac{3}{2} - \dfrac{1}{n} - \dfrac{1}{n+1}. \end{align*}$$
2. Since the terms $\dfrac{1}{n}$ and $\dfrac{1}{n+1}$ go to zero as $n$ goes to infinity, the series has a limiting sum of $\dfrac{3}{2}$.