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$n$ is a positive integer, then

$$1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac53.$$

please don't refer to the famous $1+\frac1{2^2}+\frac1{3^2}+\dotsb=\frac{\pi^2}6$.

I want to find a better proof.

My stupid method:

$$1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt \left(1+\frac1{2^2}+\dotsb+\frac1{10^2}\right)+\frac1{10\cdot11}+\dotsb+\frac1{n(n-1)}\\<1.549768...+\frac1{10}\lt\frac53$$

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  • $\begingroup$ how does it get 1k views? $\endgroup$ – Luis Felipe Oct 13 '15 at 19:06
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    $\begingroup$ Your method is just fine. It is correct (assuming your computation is correct). You were not asked for a proof from the book, just a proof from your book. $\endgroup$ – marty cohen Oct 13 '15 at 19:12
  • $\begingroup$ What do you mean by "better"? $\endgroup$ – user285523 Nov 8 '15 at 5:05
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I suggest

$$\sum_{k=1}^n \frac{1}{k^2} \leqslant 1 + \sum_{k=2}^n \frac{1}{k^2 - \frac14} = 1 + \sum_{k=2}^n \left(\frac{1}{k-\frac12} - \frac{1}{k+\frac12}\right) = 1 + \frac23 - \frac{1}{n+\frac12}.$$

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  • $\begingroup$ How did you get from $1 + \sum_{k=2}^n \left(\frac{1}{k-\frac12} - \frac{1}{k+\frac12}\right)$ to $1 + \frac23 - \frac{1}{n+\frac12}.$? $\endgroup$ – Ovi Jun 12 '16 at 4:15
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    $\begingroup$ @Ovi Telescoping: $$1 + \sum_{k = 2}^n \biggl(\frac{1}{k-\frac{1}{2}} - \frac{1}{k + \frac{1}{2}}\biggr) = 1 + \sum_{k = 2}^n \biggl(\frac{1}{k-\frac{1}{2}} - \frac{1}{(k+1)-\frac{1}{2}}\biggr) = 1 + \frac{1}{2-\frac{1}{2}} - \frac{1}{(n+1)-\frac{1}{2}}.$$ $\endgroup$ – Daniel Fischer Jun 12 '16 at 9:32
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We show by induction that if $n\gt 1$ then $$1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{1}{n^2}\lt \frac{5}{3}-\frac{2}{2n+1}.$$ The result is true at $n=2$. Suppose that the result holds at $n=k$. We show it holds at $n=k+1$.

By the induction assumption, $$1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots +\frac{1}{k^2}+\frac{1}{(k+1)^2}\lt \frac{5}{3}-\frac{2}{2k+1}+\frac{1}{(k+1)^2}.\tag{1}$$ The right-hand side of (1) is equal to $$\frac{5}{3}-\left(\frac{2}{2k+1}-\frac{1}{(k+1)^2}\right).$$ Now we need to show that $\frac{2}{2k+1}-\frac{1}{(k+1)^2}\gt \frac{1}{2k+3}$ or equivalently that $\frac{4}{(2k+1)(2k+3)}\gt \frac{1}{(k+1)^2}$. So we show that $4(k+1)^2\gt (2k+1)(2k+3)$. This is straightforward.

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  • $\begingroup$ Can I ask how that term ($\frac{2}{2n+1}$) reached to your mind in order to strength your Induction hypothesis ? $\endgroup$ – Fardad Pouran Jun 7 '14 at 13:01
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    $\begingroup$ Nothing as nice as the (identical) choice by Daniel Fischer. A glance at the series shows how the error term behaves, so I looked for a suitable $\frac{1}{2n+ a}$. I think! It was a while ago. $\endgroup$ – André Nicolas Jun 7 '14 at 14:39
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I don't know if this is "better," but since $1/x^2$ is strictly decreasing, we have

$$\sum_{n=N+1}^\infty{1\over n^2}\lt\int_N^\infty{1\over x^2}\,dx={1\over N}$$

after which a little systematic trial and error gives

$$\sum_{n=1}^\infty{1\over n^2}\lt 1+{1\over4}+{1\over9}+{1\over16}+{1\over25}+{1\over5}={5989\over3600}\lt{6000\over3600}={5\over3}$$

Added later: It occurs to me that my approach is not inherently different from the OP's "stupid" method. We each basically argue that

$$\sum_{n=1}^\infty{1\over n^2}\lt1+{1\over4}+\cdots+{1\over N^2}+{1\over N}$$

for any $N$. If there's anything stupid in the OP's solution (which there isn't, really), it's just that it's not necessary to go all the way out to $N=10$.

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