# Math Help - Bounds on sum of reciprocal of squares

1. ## Bounds on sum of reciprocal of squares

I would like to have a bound on the on the following sum which for me will do but am not able to prove it. Can someone help please?

Is it true: $\forall m>1$, we have

$\sum_{n>m}\frac{1}{n^2}<\frac{1}{m-1}$?

2. Originally Posted by Abegunde
I would like to have a bound on the on the following sum which for me will do but am not able to prove it. Can someone help please?

Is it true: $\forall m>1$, we have

$\sum_{n>m}\frac{1}{n^2}<\frac{1}{m-1}$?

$\displaystyle{\sum\limits^\infty_{n>m}\frac{1}{n^2 }\leq \sum\limits^\infty_{n>m}\frac{1}{n(n-1)}=\sum^\infty_{n>m}\left(\frac{1}{n-1}-\frac{1}{n}\right)$ ....

Tonio