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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?

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  • $\begingroup$ Thank you everybody, got it! =) $\endgroup$ – jjepsuomi Sep 4 '17 at 14:19
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$$\begin{array}{rcl} \displaystyle \sum_{n=m+1}^\infty \frac{1}{n^2} &\le& \displaystyle \sum_{n=m+1}^\infty \frac{1}{(n-1)n} \\ &=& \displaystyle \sum_{n=m+1}^\infty \left(\frac{1}{n-1} - \frac1n \right) \\ &=& \displaystyle \left(\frac{1}m - \frac1{m+1} \right) + \left(\frac{1}{m+1} - \frac1{m+2} \right) + \cdots \\ &=& \dfrac1m \end{array}$$

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Hint:

For each $n$ you have $$\frac1{n^2}\le\frac1{n(n-1)}=\frac1{n-1}-\frac1n.$$

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Hint: Compare with the integral $\int_{m+1}^{\infty} {1\over x^2}dx$.

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  • $\begingroup$ You probably want the lower limit to be $1/m$ for an upper bound $\endgroup$ – Henry Sep 4 '17 at 14:17

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