How to prove that this sequence is not a Cauchy sequence

Question

I'm trying to prove that $\{a_n\}_{n\in\mathbb{N}}$ is not a Cauchy sequence, where $$a_n=\sum_{k=1}^n\frac{1}{k}$$ I guess I should use the definition of Cauchy sequence and write $$|a_m-a_n|=|a_{n+p}-a_n|=\left|\sum_{k=n+1}^{n+p}\frac{1}{k}\ \right|=\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{n+p}$$ But I don't know how to find an upper bound for this sum.

Answer 1

Definition of not Cauchy: There exists $\epsilon>0$ such that for all $N$ there exists $n,m\geq N$ with $|a_n-a_m|\geq \epsilon$.

For any $N$, consider $|a_{2N}-a_N|=\sum_{k=N}^{2N}\frac1k=\underbrace{\frac1N+\frac{1}{N+1}+\dots +\frac{1}{2N}}_{N \,\text{terms}}\geq N\cdot\frac{1}{2N}=\frac12.$ Hence, choose $\epsilon=\frac12$.

Answer 2

HINT: A sequence $x_n$ is Cauchy if $(\forall \varepsilon >0)$ $(\exists N)$ $$|x_m-x_n|< \varepsilon$$ $(\forall n,m \geq N)$. So you need to show the negation of this statement, I.E. $(\exists \varepsilon >0)$ $(\forall N)$ $$|x_m-x_n|\geq \varepsilon$$ for some $m,n \geq N$. This follows in a similar manner of the proof of divergence of the harmonic series.