How do I finish this proof proving the divergence of the harmonic series?

Question

For a project I have to prove the divergence of $\zeta(1)$, more commonly known as the Harmonic series. I'm having trouble with this because I once managed to do it but I left the paper that I used somewhere and I'm unable to remember where. Here's the part that I remember: $$H_\infty = \sum_{k=1}^{\infty} (\frac{1}{k})$$ Parts of the sum can be written as $$P_n = \sum_{k=n}^{2n} (\frac{1}{k})$$ and $$H_\infty = 1 + P_2 + P_5 + P_{11} + ...$$ We can calculate $\frac{1}{n} + \frac{1}{2n} = 1\frac{1}{2} * \frac{1}{n}$ and we know that $\frac{1}{n + p} + \frac{1}{2n-p} < \frac{1}{n} + \frac{1}{2n}$ because $\frac{1}{n} - \frac{1}{n+p} > \frac{1}{2n-p} - \frac{1}{2n}$. Now I don't know how to continue. Since $\frac{1}{n} + \frac{1}{2n} = 1\frac{1}{2} * \frac{1}{n}$, we can say that $\sum_{k=n}^{2n} (\frac{1}{k}) < \frac{1}{2}n * 1\frac{1}{2} * \frac{1}{n}$ thus $P_n < \frac{3}{4}$. I don't have the feeling, however, that this helps me any further as this only proves it's smaller that a certain value, not bigger, as I would've hoped. Can anyone help me out?

Answer 1

The argument can be done as simple as this: All the summands in $P_n$ are larger or equal to $\tfrac{1}{2n}$ and there are $n+1$ summands, thus

$P_n = \sum_{k=n}^{2n} \frac{1}{k} \geq \sum_{k=n}^{2n} \frac{1}{2n} = (n+1) \frac{1}{2n} \geq \frac{1}{2}.$

Answer 2

Usual proofs exploit the Cauchy condensation test or the integral test, but you may use that $\log(1+x)<x$ for any $x>0$ follows by convexity, hence $$ H_N = \sum_{n=1}^{N}\frac{1}{n} > \log\prod_{n=1}^{N}\left(1+\frac{1}{n}\right)\stackrel{\text{telescopic!}}{=}\log(N+1) $$ and the RHS is clearly divergent as $N\to +\infty$.