- #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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