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 $$\sum_{n=1}^\infty \frac 1n = \infty\text.$$

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.