Prove that sequence it's a Cauchy Sequence ( $\mid a_{n+1}-a_n \mid\leq \frac{1}{n^2}$ )

Question

A friend who just started calculus asked me to help him on this question: Prove that a sequence {$a_n$}, which verifies the following inequality, converge : \begin{equation} \mid a_{n+1}-a_n\mid \leq \frac{1}{n^2} \end{equation}

My try it's to show it's a Cauchy sequence, for prove it I do this:

Assume $m>n$ for $m,n$ enough large, then: \begin{align} \mid a_m-a_n\mid\; &=\; \mid a_m-a_{m-1}\;+a_{m-1}\;-a_{m+2}\; +a_{m+2}\;\;...\;+\; a_{n+1}\,-a_n\mid\ \\ &\leq \; \mid a_m-a_{m-1}\;\mid+\mid a_{m-1}\; -a_{m+2}\; \mid +\;...\; +\mid a_{n+1}\,-a_n\;\mid\ \\ &\leq \frac{1}{(m-1)^2}+\frac{1}{(m-2)^2}+\;...+\frac{1}{n^2}\; = \sum_{k=n}^{m-1}\frac{1}{k^2} \\ & \end{align}

And I get stucked there because I obviously know that this sum converges, and then the last terms of the sum must decay a lot, so they must be lower than a positive epsylon. But he didn't learn series yet, so I don't know if I'm taking the exercise by the right path.

Thanks!

Answer 1

If you don't know series, use this: $$ \frac{1}{n^2} < \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n} . $$ Can you make your argument work with that?