Harmonic Series is Divergent

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Theorem

The harmonic series:

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

diverges.


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.

$\blacksquare$


Proof 5

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.

$\blacksquare$


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