Proof for a sum to be less than an integral of the sums expression.

1. Feb 1, 2009

Gregg

1. The problem statement, all variables and given/known data

Prove that

$\displaystyle\sum_{x=2}^n \frac{1}{x^2}<\int_0^n \frac{1}{x^2} dx$

2. Relevant equations

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

$\displaystyle\sum_{x}^{\infty} \frac{1}{x^s} = \displaystyle\prod_{p}^{\infty} \frac{1}{1-p^{-s}}$

Or maybe

$\displaystyle\sum_{x}^{\infty} \frac{1}{x^2} = \frac{\pi^2}{6}$

3. The attempt at a solution

(1)

$\displaystyle\sum_{x=2}^{\infty} \frac{1}{x^2} = \displaystyle\sum_{x}^{\infty} \frac{1}{x^2} - 1$

$\displaystyle\sum_{x=2}^{\infty} \frac{1}{x^2} = \frac{\pi^2}{6} - 1$

(2)

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

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

This integral does not converge between those limits?

(3)

$\frac{\pi^2}{6} - 1 > 0$

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

Last edited: Feb 1, 2009
2. Feb 1, 2009

Dick

There's a much easier approach. Show a lower Riemann sum for the integral that equals the series. Are you sure you want the lower limit on the integral to be 0? That's divergent, you know. I'm guessing it should be 1.

3. Feb 1, 2009

Gregg

Yes it was an error the lower limit is not 0 it is 1.

(1)

$\displaystyle\sum_{x=2}^{\infty} \frac{1}{x^2} = \displaystyle\sum_{x}^{\infty} \frac{1}{x^2} - 1$

$\displaystyle\sum_{x=2}^{\infty} \frac{1}{x^2} = \frac{\pi^2}{6} - 1$

(2)

$\int_1^\infty \frac{1}{x^2} dx = \left[ -\frac{1}{x} \right]_1^\infty$

$\left[ -\frac{1}{x} \right]_1^\infty = -\frac{1}{\infty} + 1 = 1$ ?

(3)

$\frac{\pi^2}{6} - 1 < 1$

It is proved for infinity now I think but is this enough to say that it is so for

$\displaystyle\sum_{x=2}^n \frac{1}{x^2} < \int_1^n \frac{1}{x^2} dx$ ?

4. Feb 1, 2009

Dick

Yes, the integral is 1 and pi^2/6-1 is less than one. That's fine. But what I'm suggesting is a more low tech solution in which you don't need to know that the sum of the inverse squares is pi/6. Draw the graph of 1/x^2 and fit rectangles under the curve between every pair of integers. The sum of all the rectangles is clearly less than the integral, right?

5. Feb 1, 2009

Gregg

Oh... I've just realised that the rectangles you're talking about actually fit that sum in the question!