### Infinite series: Common series

### The harmonic series

In this section we look at the sequence defined by \(s_k = \sum_{n=1}^k \frac{1}{n}\) . This means that \[ \begin{aligned} s_1 &= 1 \\ s_2 &= 1 + \frac{1}{2} \\ s_3 &= 1 + \frac{1}{2} + \frac{1}{3}\text, \end{aligned} \] etc. The terms of this sum, \(\frac{1}{n}\) , converge to zero as \(n\to \infty\), but we will see that \(\lim_{k\to\infty} s_k = \infty\) , i.e. \[ \sum_{n=1}^\infty \frac{1}{n} = \infty\text. \] This series is called the harmonic series.

We group the terms in the infinite sum in the following way: \[ 1 + \frac{1}{2} + \underbrace{\frac{1}{3} + \frac{1}{4}}_{\text{first block}} + \underbrace{\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}}_{\text{second block}} + \underbrace{\frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + \frac{1}{13} + \frac{1}{14} + \frac{1}{15} + \frac{1}{16}}_{\text{third block}} + \dots \] We leave the \(1\) and \(\tfrac 12\) and then create blocks such that each next block is twice as big as the previous one. Since \(\tfrac 1n\) is decreasing the sum of successive terms is less than the last term multiplied by the number of elements in a block. For example, for the first block we have \[ \frac 13 + \frac 14 > 2 \cdot \frac 14 = \frac 12\] and similarly for the second block \[ \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}> 4 \cdot \frac 18 = \frac 12\text. \] We see that the sum of the terms in a block is always at least \(\tfrac 12\). \[ 1 + \frac{1}{2} + \underbrace{\frac{1}{3} + \frac{1}{4}}_{\gt \tfrac 12} + \underbrace{\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}}_{\gt \tfrac 12} + \underbrace{\frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + \frac{1}{13} + \frac{1}{14} + \frac{1}{15} + \frac{1}{16}}_{\gt\tfrac 12} + \dots \] For each \(M > 0\) there exists a \(K\) such that \(s_k > M\) for all \(k > K\) . This means that \(\lim_{k\to\infty} s_k = \infty\) , or \[ \sum_{n=1}^\infty \frac 1n = \infty\text. \]

Warning Please note that even though a sequence converges to zero, the series does not necessarily converge. The harmonic series \[ \sum_{n=1}^\infty \frac 1n = \infty \] illustrates this.

More general, the series \[ \sum_{n=1}^\infty \frac{1}{n^p} \] converges precisely if \(p > 1\). We will prove this in the theory page entitled The integral test.