Question

Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+2}\right) $$

Short answer

The given series converges and the sum of the series is \(3/2\).

Step-by-step solution

Split the Series

The first step is to split the given series into two simpler series. Given series is: \[\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+2}\right)\] This can be written as: \[\sum_{n=1}^{\infty}\frac{1}{n} - \sum_{n=1}^{\infty}\frac{1}{n+2}\]

Shift the Index of the Second Series

The second step is to shift the index of the second series as it starts from \(n+2\). So, let \(n+2 = m\). When \(n = 1\), \(m\) starts from \(3\). Then, \(\sum_{n=1}^{\infty}\frac{1}{n+2}\) becomes \(\sum_{m=3}^{\infty}\frac{1}{m}\)

Find the Difference of the Two Series

Subtract the second series from the first one, \[\sum_{n=1}^{\infty}\frac{1}{n} - \sum_{m=3}^{\infty}\frac{1}{m} = \frac{1}{1} + \frac{1}{2} - \sum_{m=2}^{\infty}\frac{1}{m}\]This simplifies to \(1 + 1/2 - 0 = 3/2\)

Conclude the Convergence/Divergence

Since the infinite series simplifies to a finite constant, we can conclude that the series converges and its limit is that finite constant.