# Harmonic Series is Divergent

## Theorem

The harmonic series:

- $\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 n$

## Proof 1

- $\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 n = \underbrace 1_{s_0} + \underbrace {\frac 1 2 + \frac 1 3}_{s_1} + \underbrace {\frac 1 4 + \frac 1 5 + \frac 1 6 + \frac 1 7}_{s_2} + \cdots$

where $\displaystyle s_k = \sum_{i \mathop = 2^k}^{2^{k + 1} \mathop - 1} \frac 1 i$

From Ordering of Reciprocals:

- $\forall m, n \in \N_{>0}: m < n: \dfrac 1 m > \dfrac 1 n$

so each of the summands in a given $s_k$ is greater than $\dfrac 1 {2^{k + 1} }$.

The number of summands in a given $s_k$ is $2^{k + 1} - 2^k = 2 \times 2^k - 2^k = 2^k$, and so:

- $s_k > \dfrac{2^k} {2^{k + 1} } = \dfrac 1 2$

Hence the harmonic sum $H_{2^m}$ satisfies the following inequality:

\(\displaystyle H_{2^m}\) | \(=\) | \(\displaystyle \sum_{n \mathop = 1}^{2^m} \frac 1 n\) | |||||||||||

\(\displaystyle \) | \(>\) | \(\displaystyle \sum_{n \mathop = 1}^{2^m - 1} \frac 1 n\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \sum_{k \mathop = 0}^m \left({s_k}\right)\) | |||||||||||

\(\displaystyle \) | \(>\) | \(\displaystyle 1 + \sum_{a \mathop = 0}^m \frac 1 2\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1 + \frac m 2\) |

The right hand side diverges, from the $n$th term test.

The result follows from the the Comparison Test for Divergence.

$\blacksquare$

## Proof 2

Observe that all the terms of the harmonic series are strictly positive.

From Reciprocal Sequence is Strictly Decreasing, the terms are decreasing.

Hence the Cauchy Condensation Test can be applied, and we examine the convergence of:

\(\displaystyle \sum_{n \mathop = 1}^\infty 2^n \frac 1 {2^n}\) | \(=\) | \(\displaystyle \sum_{n \mathop = 1}^\infty 1\) |

This diverges, from the $n$th term test.

Hence $\displaystyle \sum \frac 1 n$ also diverges.

$\blacksquare$

## Proof 3

We have that the Integral of Reciprocal is Divergent.

Hence from the Integral Test, the harmonic series also diverges.

$\blacksquare$

## Proof 4

For all $N \in \N$:

- $\dfrac 1 N + \dfrac 1 {N + 1} + \cdots + \dfrac 1 {2 N} > N \cdot \dfrac 1 {2 N} = \dfrac 1 2$

Hence, by Cauchy's Convergence Criterion for Series, the Harmonic series is divergent.

$\blacksquare$

## Also see

## Historical Note

The proof that the Harmonic Series is Divergent was discovered by Nicole Oresme.

However, it was lost for centuries, before being rediscovered by Pietro Mengoli in $1647$.

It was discovered yet again in $1687$ by Johann Bernoulli, and a short time after that by Jakob II Bernoulli, after whom it is usually (erroneously) attributed.

Some sources attribute its rediscovery to Jacob Bernoulli.

## Sources

- 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $23 \cdotp 103 \, 45 \ldots$ - 1992: Larry C. Andrews:
*Special Functions of Mathematics for Engineers*... (previous) ... (next): $\S 1.2.2$: Summary of convergence tests - 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {B}.19$: The Series $\sum 1/ p_n$ of the Reciprocals of the Primes - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $23 \cdotp 10345 \ldots$