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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$$\sum_{k=2}^{n}\frac{1}{k^2}\leq \sum_{k=2}^{n}\frac{1}{k^2-k}$$ The Series in RHS is telescoping.
Fermat
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2You might want to be more careful there... RHS is not defined for $k = 1$.. – Ivo Terek Feb 23 '15 at 20:51
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@ivo terek do you see $k=2$ ? – N.S.JOHN Mar 28 '16 at 12:32
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This comment is a year old now. He probably corrected the $k=1$ immediately after I pointed it. – Ivo Terek Mar 28 '16 at 12:34
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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.
Olivier Oloa
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