# How to prove $\displaystyle{\sum_{n=m+1}^\infty} \frac{1}{n^2}\leq \frac1m$

I need to prove the following inequality:

$$\sum_{n=m+1}^\infty \frac{1}{n^2}\leq \frac1m$$

But I'm stuck with it. I found online geometric justifications for this but I'd really appreciate to see actual proof. Any hints?

• Thank you everybody, got it! =) – jjepsuomi Sep 4 '17 at 14:19

$$\begin{array}{rcl} \displaystyle \sum_{n=m+1}^\infty \frac{1}{n^2} &\le& \displaystyle \sum_{n=m+1}^\infty \frac{1}{(n-1)n} \\ &=& \displaystyle \sum_{n=m+1}^\infty \left(\frac{1}{n-1} - \frac1n \right) \\ &=& \displaystyle \left(\frac{1}m - \frac1{m+1} \right) + \left(\frac{1}{m+1} - \frac1{m+2} \right) + \cdots \\ &=& \dfrac1m \end{array}$$
For each $n$ you have $$\frac1{n^2}\le\frac1{n(n-1)}=\frac1{n-1}-\frac1n.$$
Hint: Compare with the integral $\int_{m+1}^{\infty} {1\over x^2}dx$.
• You probably want the lower limit to be $1/m$ for an upper bound – Henry Sep 4 '17 at 14:17