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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_na_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_ns_m\geq \epsilon##
Set ##\epsilon=\frac{1}{2}##. Let ##N\in \mathbb{N}## and choose ##n=N,m=2N##. Then:
##\begin{align*}
\lefts_Ns_{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*}##
##\begin{align*}
\lefts_Ns_{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*}##