How to prove divergence of harmonic series by eps-delta proof?

[Eclair_de_XII]

Homework Statement

Let $\{s_n\}_{n\in \mathbb{N}}=\{\sum_{1\leq k \leq n} \frac{1}{k}\}_{n\in \mathbb{N}}$. Show that this sequence cannot converge.

Relevant Equations

A sequence $\{a_n\}_{n\in \mathbb{N}}$ is Cauchy (or equivalently, convergent) iff for all $\epsilon >0$, there is a positive integer $N$ such that for all $n,m\geq N$, $|a_n-a_m|<\epsilon$.

Want to prove:
There is a $\epsilon>0$ such that for all positive integers $N$, there is $n,m\geq N$ such that $|s_n-s_m|\geq \epsilon$

Set $\epsilon=\frac{1}{2}$. Let $N\in \mathbb{N}$ and choose $n=N,m=2N$. Then:

$\begin{align*} \left|s_N-s_{2N}\right|&=&\left|\sum_{l=1}^N \frac{1}{l} - \sum_{l=1}^{2N} \frac{1}{l}\right|\\ &=&\left|\left(1+\frac{1}{2}+...+\frac{1}{N}\right)-\left(1+...+\frac{1}{N}+...+\frac{1}{2N}\right)\right|\\ &=&\left|\frac{1}{N+1}+...+\frac{1}{N+N}\right|\\ &>&\left|\frac{1}{N+N}+...+\frac{1}{N+N}\right|=\left|\frac{N}{N+N}\right|\\ &=&\left|\frac{N}{2N}\right|=\frac{1}{2}=\epsilon \end{align*}$

 

[Delta2]

I think your work is correct. You prove that the sequence of partial sums is not Cauchy, and hence (by equivalence of the notions of Cauchy and Convergence for a sequence) is not convergent.

 

[Eclair_de_XII]

Thanks for checking my work.