Show that $\sum^\infty_{n=m}\frac{1}{n^2} = O(\frac{1}{m})$

Question

I have been told it is very simple, but I cannot work it out. I was thinking I use that I know the left hand side converges?

Answer 1

If $m\geq 2$, then $$ \sum\limits_{n = m}^\infty {\frac{1}{{n^2 }}} \le \sum\limits_{n = m}^\infty {\frac{1}{{n(n - 1)}}} = \sum\limits_{n = m}^\infty {\left( {\frac{1}{{n - 1}} - \frac{1}{n}} \right)} = \frac{1}{{m - 1}} \leq \frac{2}{m}. $$