# Prove that $\sum_n \frac{1}{n^2} < 2$

1. Jul 16, 2016

### Portuga

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

Prove that $$1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots+\frac{1}{n^{2}}<2.$$

2. Relevant equations

$$1 = 2^0, 2 = 2^1, 3 = 2^2 - 1, 4 = 2^2, \ldots,$$

and

$$2^{n}<2^{n}+1<\ldots<2^{n+1}-1$$

3. The attempt at a solution
As $$2^{n}<2^{n}+1<\ldots<2^{n+1}-1,$$ it's true that
$$\frac{1}{2^{n}}>\frac{1}{2^{n}+1}>\ldots\frac{1}{2^{n+1}-1}.$$
So, being
$$S_{n}=\left(\frac{1}{2^{n}}\right)\left(\frac{1}{2^{n}}\right)+\left(\frac{1}{2^{n}+1}\right)\left(\frac{1}{2^{n}+1}\right)+\ldots+\left(\frac{1}{2^{n+1}-1}\right)\left(\frac{1}{2^{n+1}-1}\right),$$
follows from previous inequalities that

$$S_{n} < \left(\frac{1}{2^{n}}\right)\left(\frac{1}{2^{n}}\right)+\ldots+\left(\frac{1}{2^{n}}\right)\left(\frac{1}{2^{n}}\right)$$

$$=\frac{n}{2^{2n}}.$$

This is a dead-end to me, because I can't compare it with 2 in an elegant way. I made the graphic of this function, $$f(x) = x/2^{2x},$$ and I verified it's less than 2 for all x, but I am pretty sure there is a better way to do this proof.

2. Jul 16, 2016

### Math_QED

Are you familiar with induction?

3. Jul 16, 2016

### RedDelicious

Assume that it is greater than or equal 2 and then show that can't be true

$$\sum \frac{1}{n^2}=\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\ge 2 \\ \frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\ge 1 \\ \frac{2^2+3^2+4^2+...+n^2}{2^23^24^2...n^2}\ge 1 \\$$

which you can show isn't true using a few different arguments

4. Jul 16, 2016

### Ray Vickson

$$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} = \frac{2^2+3^2+4^2+...+n^2}{2^23^24^2...n^2} \:\Longleftarrow\:\text{FALSE!}$$

5. Jul 16, 2016

### Portuga

Yes, I will try this strategy.

6. Jul 16, 2016

### Portuga

Thank you very much!

7. Jul 16, 2016

### haruspex

Your Sn only sums the 1/r2 for r from n to 2n. You need the sum all the way from 1.

With regard to induction,you would need to set up your inductive hypothesis cleverly. Starting with "suppose the sum of the first N terms < 2" will get you nowhere. You'd need to find some other series summing to at most 2 and show that it acts as an upper bound for the given sequence.

A useful strategy sometimes is to compare the series with the corresponding continuous function.

Last edited: Jul 16, 2016
8. Jul 16, 2016

### Portuga

Thanks. I will find one.

9. Jul 16, 2016

### rcgldr

I see an issue with the first of the relevant equations, based on the goal of this problem, the first relevant equation should be:

$$1 = 2^0, \ 2 = 2^1, 3 = 2^1+1, \ 4 = 2^2, 5 = 2^2+1, 6 = 2^2+2, 7 = 2^2+3, 8 = 2^3, \ \ldots$$

I don't think mathematical induction will be part of the solution. I'll wait for a reply before providing any more hints.

Last edited: Jul 18, 2016
10. Jul 17, 2016

### ehild

Yes, it is an excellent hint!
Compare the sum to the definite integral of 1/x2 from 1 to n.

11. Jul 17, 2016

### haruspex

I was avoiding being too specific at this stage.

12. Jul 17, 2016

### ehild

Your hint "compare the series with the corresponding continuous function" was a bit misleading. I think you meant the series Σ1/n2 compared to the integral of the corresponding continuous function. With the figure, I wanted to make your hint clear .

13. Jul 17, 2016

### haruspex

Again, I was leaving it to the OP to make the analogy between the sum of the series and the integral of the function.

14. Jul 17, 2016

### ehild

Series is the sum of a sequence. https://en.wikipedia.org/wiki/Series_(mathematics). I think you meant the sum of the sequence 1/n2.

15. Jul 17, 2016

### haruspex

16. Jul 20, 2016

### rcgldr

Combining the attempt in post #1 and the relevant equation in post #9:

$$1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\ldots < 1+\left(\frac{1}{2^{2}}+\frac{1}{2^{2}}\right)+\left(\frac{1}{4^{2}}+\frac{1}{4^{2}}+\frac{1}{4^{2}}+\frac{1}{4^{2}}\right)+ \ldots$$

A finite version of the right hand series is equal to:

$$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{N}$$

Multiply by 2

$$2+\frac{2}{2}+\frac{2}{4}+\frac{2}{4}+\ldots+\frac{2}{N} = 3+\frac{1}{2}+\frac{1}{4}+\ldots+\frac{2}{N}$$

Then subtract 2 x series - 1 x series to eliminate the inner terms:

$$\left(3+\frac{1}{2}+\frac{1}{4}+\ldots+\frac{2}{N}\right) - \left(1+\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{N}\right) = 2 - \frac{1}{N}$$

Since

$$1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\ldots < 1+\left(\frac{1}{2^{2}}+\frac{1}{2^{2}}\right)+\left(\frac{1}{4^{2}}+\frac{1}{4^{2}}+\frac{1}{4^{2}}+\frac{1}{4^{2}}\right)+ \ldots$$

Then even an infinite series of the left hand side is < 2.

Last edited: Jul 20, 2016
17. Jul 20, 2016

### Ray Vickson

You could also use the fact that for $n \geq 2$ we have
$$\frac{1}{n^2} < \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$$
The series
$$\sum_{n=2}^N \left( \frac{1}{n-1} - \frac{1}{n} \right)$$
is easy to deal with (being a so-called "telescoping" series).