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 \(\dfrac{3}{2}\).