Proof of summation 1/i^2 <= 2 - 1/n

**Bruins1** · September 17th 2009, 09:13 AM

I have put up an attachment .doc file with my work. I would like to know if my proof is correct.

Thanks

**Krizalid** · September 17th 2009, 12:43 PM

there is a short way to prove this by using some of calculus.

are you allowed to use some of calculus?

**PaulRS** · September 17th 2009, 01:54 PM

Originally Posted by Bruins1

I have put up an attachment .doc file with my work. I would like to know if my proof is correct.

Thanks

It's not correct. You say that $\sum_{k=1}^{n}{\tfrac{1}{k^2}}=\tfrac{\pi^2}{6}$ when that is not so, actually: $\sum_{k=1}^{+\infty}{\tfrac{1}{k^2}}=\tfrac{\pi^2} {6}$ , besides you use approximations and not the real value of the sum.

Note that: $\tfrac{1}{k^2}<\tfrac{1}{k\cdot (k-1)}=\tfrac{1}{k-1}-\tfrac{1}{k}$ for $k>1$

What do you see if you sum on both sides from $k=2$ to $k=n$ ? - add afterwards any missing terms-

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