Why the harmonic series diverges

Aestusy

One of the first beautiful surprises in analysis is that the harmonic series diverges, even though its terms go to zero.

The series is

$$
1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots
$$

At first it feels like it should converge, because the terms (1/n) become smaller and smaller. But the condition (a_n \to 0) is necessary for convergence, not sufficient.

The classic grouping argument is very neat:

$$
1 + \frac{1}{2}

\left(\frac{1}{3} + \frac{1}{4}\right)
\left(\frac{1}{5} + \cdots + \frac{1}{8}\right)
\left(\frac{1}{9} + \cdots + \frac{1}{16}\right)
\cdots
$$

Now look at each group.

For the group

$$
\frac{1}{3} + \frac{1}{4},
$$

both terms are at least (1/4), so

$$
\frac{1}{3} + \frac{1}{4} \geq \frac{1}{4} + \frac{1}{4} = \frac{1}{2}.
$$

For the next group,

$$
\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8},
$$

all four terms are at least (1/8), so

$$
\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}
\geq 4 \cdot \frac{1}{8}
= \frac{1}{2}.
$$

In general, the group

$$
\frac{1}{2^k+1} + \frac{1}{2^k+2} + \cdots + \frac{1}{2^{k+1}}
$$

contains (2^k) terms, and each term is at least (1/2^{k+1}). Therefore the whole group is at least

$$
2^k \cdot \frac{1}{2^{k+1}} = \frac{1}{2}.
$$

So the harmonic series is bounded below by

$$
1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \cdots
$$

which clearly diverges.

Therefore,

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

The moral is: terms going to zero is not enough. The terms must go to zero fast enough.