# Harmonic Series is Divergent/Proof 5

This article has been proposed for deletion. In particular: This is effectively the same proof, in overall structure, as Harmonic Series is Divergent/Proof 1, just a very roundabout way of proving it. |

## Theorem

The harmonic series:

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

## Proof

Assume that for $G \ge 4$ that $\displaystyle \sum_{n = G}^\infty \frac{1}{n} = L < \infty$. Namely, that for some number $G$, there is a tail of the harmonic series which converges.

Then from Definition:Series/Sequence of Partial Sums:

$s_N := \sum_{n = G}^N \frac{1}{n}$ is the partial sum of the above series. Which yields the sequence $\{ s_N \}$ of partial sums.

And, from Definition:Convergent Series we have that $\displaystyle \sum_{n = G}^\infty \frac{1}{n}$ converges iff $\{ s_N\}$ converges.

From Combination Theorem for Sequences/Normed Division Ring/Constant Rule : The constant sequence $\{ G \}$ has limit $G$. Note: $\R$ is a normed division ring as it is a field.

By Combination Theorem for Sequences/Real/Product Rule: The product of the sequences $\{ G \}$ and $\{ s_N\}$ has limit $GL$. Namely, the sequence $\{Gs_N\}$ has limit $GL$, by the opening assumption.

$GL = \displaystyle \sum_{n = G}^\infty \frac{G}{n} = \underbrace {1}_{s_0} + \underbrace{ \frac{G}{G+1} + \ldots + \frac{G}{G+4}}_{s_1} + \underbrace { \frac{G}{G+5} + \ldots + \frac{G}{G+12} }_{s_2} + \underbrace { \frac{G}{G+13} + \ldots + \frac{G}{G+28} }_{s_3} + \ldots $

Where $s_0 = 1, s_1 = \frac{G}{G+1} + \ldots + \frac{G}{G+4}$ and for $k \ge 2, s_k = \displaystyle \sum_{i = 2^k + 2^{k-1} + \ldots 2^2 + 1}^{2^{k+1} + 2^{k} + \ldots 2^2} \frac{G}{G+i} $.

From the above, $s_0 = 1, s_1 \ge \frac{4G}{G+4} \ge 1$ by inspection.

And for $k \ge 2$ then $s_k \ge \frac{2^{k+1}G}{G+2^{k+2}}$ since $\frac{G}{G+2^{k+2}}$ is smaller than the smallest summand of $s_k$. If summed $2^{k+1}$ many times, $2^{k+1}$ being the number of summands in $s_k$, it yields a result less than $s_k$. Note: The smallest summand of $s_k$ is $\frac{G}{G+2^{k+1}+ \ldots + 2^2}$.

Claim:

For $k \ge 1, G \ge 4$ We have $\frac{2^{k+1}G}{G+2^{k+2}} \ge 1$

Proof:

$\frac{2^{k+1}G}{G+2^{k+2}} \ge 1 \iff k + 1 \ge \log_2\frac{G}{G-2}$. If $G = 4$ then the righthand side of the second inequality is $\log_2(2) = 1$. If $G > 4$, then $ 1 < \frac{G}{G-2} < 2$. Namely as $G \uparrow$ we have $\frac{G}{G-2} \to 1$ meaning $\log_2\frac{G}{G-2} \to 0$.

$\Box$

Now,

$\displaystyle GL = s_0 + s_1 + s_2 + s_3 + \ldots \ge 1 + \frac{4G}{G+4} + \frac{8G}{G+16 } + \frac{16G}{G+32} + \ldots \ge 1 + 1 + 1 + 1 + \ldots \to \infty > GL$

Deriving a contradiction. Hence , the series does not converge which implies the sequence $\{s_N\}$ does not converge. Therefore by Tail of Convergent Sequence : A sequence $a_n$ converges iff the sequence $a_{n+N}, N \in \N$ converges. We have the tail of the harmonic series diverges for any $G$ thus the harmonic series will diverge.

$\blacksquare$

## 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.