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

• Gregg
In summary: Thanks for pointing that out!In summary, the homework statement proves that the sum of the inverse squares is less than pi/6.
Gregg

Homework Statement

Prove that

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

Homework 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

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

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:
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.

Dick said:
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.

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

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

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?

Dick said:
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?

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

1. What is the definition of a sum?

A sum is a mathematical operation in which two or more numbers, quantities, or expressions are added together to obtain a single result.

2. What is an integral?

An integral is a mathematical concept that represents the accumulation of a quantity over an interval.

3. How do you prove that a sum is less than the integral of the sum's expression?

To prove that a sum is less than the integral of the sum's expression, you can use the comparison test or the squeeze theorem. These methods involve comparing the sum to a known integral or using other inequalities to show that the sum is bounded by the integral.

4. What is the significance of proving a sum to be less than the integral of the sum's expression?

Proving a sum to be less than the integral of the sum's expression can be useful in many areas of mathematics and science. It can help in evaluating series, finding bounds for certain functions, and determining the convergence of integrals.

5. Can you provide an example of using this proof in a real-life application?

Yes, this proof can be applied in physics when calculating the work done by a variable force. The work done is represented by the integral of the force over a given distance, and by proving that the sum of the force is less than the integral, we can show that the work done is less than the total area under the curve of the force function.

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