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This question already has an answer here:

How do you prove:

$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots + \frac{1}{n^2} < 2$$

I have some competitions in my country, so I have to prepare.

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marked as duplicate by Cameron Williams, N. F. Taussig, Ross Millikan, Dario, Jack D'Aurizio Feb 23 '15 at 21:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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By writing, for $N>1$: $$ \begin{align} \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{N^2}&<\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...+\frac{1}{N \times (N-1)}\\\\ &=(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+...+(\frac{1}{N-1}-\frac{1}{N})\\\\ &=1-\frac{1}{N} \end{align} $$ then $$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{N^2}\leq 2-\frac{1}{N}$$ giving the result.

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$$\sum_{k=2}^{n}\frac{1}{k^2}\leq \sum_{k=2}^{n}\frac{1}{k^2-k}$$ The Series in RHS is telescoping.

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    $\begingroup$ You might want to be more careful there... RHS is not defined for $k = 1$.. $\endgroup$ – Ivo Terek Feb 23 '15 at 20:51
  • $\begingroup$ @ivo terek do you see $k=2$ ? $\endgroup$ – N.S.JOHN Mar 28 '16 at 12:32
  • $\begingroup$ This comment is a year old now. He probably corrected the $k=1$ immediately after I pointed it. $\endgroup$ – Ivo Terek Mar 28 '16 at 12:34

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