Harmonic Series is Divergent

Theorem
The harmonic series:


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

diverges.

Hence the Riemann Zeta Function is undefined at $s=1$.

Proof 1

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

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

From Ordering of Reciprocals, $\forall m < n: \dfrac 1 m > \dfrac 1 n$, 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:
 * $s_k > \dfrac{2^k}{2^{1+k}} = \dfrac 1 2$

Hence the harmonic sum:
 * $\displaystyle \sum_{n=1}^\infty \frac 1 n = \sum_{k=0}^\infty \left({s_k}\right) > \sum_{a=1}^\infty \frac 1 2$

the last of which diverges, from the Nth Term Test.

The result follows from the the Comparison Test for Divergence.

Proof 2
Observe that all the terms of the harmonic series are strictly positive.

To prove that the sequence is decreasing, consider:

$f: \R_{>0} \to \R$: $x \mapsto x^{-1}$

From the Power Rule for Derivatives:

Because $-x^{-2} < 0$ for all $x$ considered, from Derivative of Monotone Function, $f$ is decreasing.

As $f\left({n}\right)$ agrees with the harmonic series for all $n$ in the domain of $\sum \frac 1 n$, we conclude from Monotonicity of Sequences that the series is decreasing.

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

This diverges, from the Nth Term Test.

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

Proof 3
We have that the Integral of Reciprocal is Divergent.

From the Integral Test, the harmonic series also diverges.

Also see

 * Harmonic numbers
 * Riemann Zeta Function