- #1

Gregg

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## Homework Statement

Prove that

[itex] \displaystyle\sum_{x=2}^n \frac{1}{x^2}<\int_0^n \frac{1}{x^2} dx [/itex]

## Homework Equations

I'm not sure what equations are relevant for this, but I think for this I would need:

[itex] \displaystyle\sum_{x}^{\infty} \frac{1}{x^s} = \displaystyle\prod_{p}^{\infty} \frac{1}{1-p^{-s}}[/itex]

Or maybe

[itex] \displaystyle\sum_{x}^{\infty} \frac{1}{x^2} = \frac{\pi^2}{6}[/itex]

## The Attempt at a Solution

(1)

[itex]\displaystyle\sum_{x=2}^{\infty} \frac{1}{x^2} = \displaystyle\sum_{x}^{\infty} \frac{1}{x^2} - 1[/itex]

[itex]\displaystyle\sum_{x=2}^{\infty} \frac{1}{x^2} = \frac{\pi^2}{6} - 1[/itex]

(2)

[itex]\int_0^\infty \frac{1}{x^2} dx = \left[ -\frac{1}{x} \right]_0^\infty[/itex]

[itex]\left[ -\frac{1}{x} \right]_0^\infty = 0[/itex] ?

This integral does not converge between those limits?

(3)

[itex]\frac{\pi^2}{6} - 1 > 0[/itex]

So it seems that I can't prove for this? Am I right in using infinity for the limits to prove for n or do I need another method? Thanks

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