I was looking through a proof that the sum of $$\sum_{i=1}^\infty \frac{1}{i^2} = 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}...$$ is convergent. I know there are probably other ways to prove this, but because I'm learning about functional analysis I want to prove it by first proving that this sequence is Cauchy.
The proof goes as follows. Consider any $\epsilon>0$, I can find you an $N$ in the sequence that is sufficiently large enough such that $\frac{1}{n}>\sqrt\frac{\epsilon}{2 }$ and $\frac{1}{m}>\sqrt\frac{\epsilon}{2 }$, where $n$ and $m$ are defined as greater than $N$. Thus:
$$|\frac{1}{n^2}-\frac{1}{m^2}|<\frac{1}{n^2}+\frac{1}{m^2}<\epsilon$$
Then the sequence is Cauchy.
But then I thought about the sequence: $$\sum_{i=1}^\infty \frac{1}{i} = \frac{1}{1}+\frac{1}{2}+\frac{1}{3} + \cdots.$$ If I just modify the proof a bit such that I let $\frac{1}{n}>\frac{\epsilon}{2}$ and likewise for $m$, then won't it prove that this sequence is Cauchy as well? But this sequence is obviously not Cauchy. What's wrong with applying the above proof to this series?
Hint:
$$\sum_{k = n+1}^m \frac{1}{k^2} < \sum_{k = n+1}^m \frac{1}{k(k-1)} = \sum_{k = n+1}^m \left(\frac{1}{k-1}- \frac{1}{k}\right) = \ldots$$