# Harmonic Series

## Key Questions

• The harmonic series diverges.
${\sum}_{n = 1}^{\infty} \frac{1}{n} = \infty$

Let us show this by the comparison test.
${\sum}_{n = 1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \cdots$
by grouping terms,
$= 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \cdots$
by replacing the terms in each group by the smallest term in the group,
$> 1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\right) + \cdots$
$= 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \cdots$
since there are infinitly many $\frac{1}{2}$'s,
$= \infty$

Since the above shows that the harmonic series is larger that the divergent series, we may conclude that the harmonic series is also divergent by the comparison test.

• Since the harmonic series is known to diverge, we can use it to compare with another series. When you use the comparison test or the limit comparison test, you might be able to use the harmonic series to compare in order to establish the divergence of the series in question.

• The harmonic series is
${\sum}_{n = 1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots$
(Note: This is a divergent series.)