Sum of 1/n²

By The Editor :

This is a merger of two threads on the same subject.

By Anthony Curtis (P684) on Tuesday, February 1, 2000 - 02:41 pm :

How can it be proved that the sum from 1 to infinity of n to the power minus two, equals pi squared divided by six?

i.e.

1 + 1 / 4 + 1 / 9 + 1 / 16 + 1 / 25 + 1 / 36 + \dots = π^{2} / 6

By Dan Goodman (Dfmg2) on Tuesday, February 1, 2000 - 04:38 pm :

Here is a web page devoted to that number:
Here

[Dan also provided a link to a file with 12 different proofs of the result. That file got deleted, but was probably very similar to this one . You should be warned that the proofs involve work at A-level and beyond. - The Editor]

By Jack Farchy (P1851) on Sunday, February 6, 2000 - 04:20 pm :

Surely if you're going to infinity, it equals infinity, not

π^{2} / 6

?
Jack

By Pras Pathmanathan (Pp233) on Monday, February 7, 2000 - 12:01 pm :

Ah but you can have a sum that involves an infinite number of terms being added together, but still "tend to" a finite number, so that the sum to infinity is finite, not infinity. By "tend to", I mean that you get closer and closer to that number as you add more and more of the terms. Try this simple example on the calculator:

$1 / 2 + 1 / 4 + 1 / 8 + 1 / 16 + 1 / 32 + 1 / 64 + 1 / 128 + \dots$

I'm sure you'll quickly see that this is getting closer and closer to 1. In fact it will get as close as we like to 1 eventually but never quite get there (and so certainly never make it to infinity). We can see this by realizing that whenever we add a new term, we get to a point halfway between where we were and 1 (try drawing a line, and then marking off points at half the length of the line, then $1 / 2 + 1 / 4$ the length of the line, then $1 / 2 + 1 / 4 + 1 / 8$ , and so on, and you should be able to see this, that each time you halve the distance to 1).

[Or look at this picture... - The Editor]

We say that

1 / 2 + 1 / 4 + 1 / 8 + \dots

tends to 1, or that the sum from

n = 1

to infinity of

2^{- n} = 1

. Note that I haven't really proved that the answer is 1, but it isn't very hard to do, and I can show you if you want.

Pras

By Brad Rodgers (P1930) on Friday, April 7, 2000 - 06:28 pm :

How would I find out what the series
1/2² + 1/3² + 1/4² ......
equals?
It seems to converge to pi/6, but how can I prove this?

By Dan Goodman (Dfmg2) on Saturday, April 8, 2000 - 02:30 am :

1 / 1^{2} + 1 / 2^{2} + 1 / 3^{2} + \dots = π^{2} / 6

is a famous result, and there are many (quite difficult) proofs of this. The following is a proof due to Euler about 1734, it's logically completely flawed, but here it is anyway.

$\sin (s)$ has zeros at $\pm n π$ . Therefore

$\sin (s) = s (1 - s^{2} / π^{2}) (1 - s^{2} / 4 π^{2}) (1 - s^{2} / 9 π^{2}) \dots$

Using the relationship between roots and coefficients of polynomials (which doesn't hold for infinite polynomials, but anyway), you can deduce the above formula.

(That was straight from Morris Kline's "Mathematical Thought: from Ancient to Modern Times" in case you're interested)

By Brad Rodgers (P1930) on Monday, April 10, 2000 - 09:19 pm :

I was able to find another proof, that, while complex, I could still follow.

Brad

By Dan Goodman (Dfmg2) on Tuesday, April 11, 2000 - 12:03 am :

Could you post your proof, or a link to it? This is a question that comes up a lot, it would be nice to have an easier answer to it than the ones above.

By Brad Rodgers (P1930) on Wednesday, April 12, 2000 - 09:47 pm :

Well, I can't exactly remember the proof, and unfortunately found it in A Barnes and Noble Bookstore that is quite far away from my home(I was on vacation). If I can remember correctly the books name was something like "Mathematic's History" or something along those lines. I didn't buy it as it cost $50+ and that was really the only thing that I wished to know from it. I do remember that it simply proved that by showing 1/6 = 1/(pi² )+ 1/(2pi)² ......
I'll work on remembering what the proof was.

Sorry,

Brad

By Alex Barnard (Agb21) on Monday, May 15, 2000 - 02:14 pm :

If you believe in the existence of Fourier series (and know how to compute them) then you'll be able to show the following:

The function f(x)=2 x Pi x|x| on (-Pi,Pi) has a Fourier series:

Pi² - 8 x (cos(x)/1² + cos(3x)/3² + cos(5x)/5² + ...)

Now stick in the value x=0 and we get:

0 = Pi² - 8(1/1² + 1/3² + 1/5² + ...)

and so:

Pi² /8 = 1/1² + 1/3² + 1/5² + ...

So, how do we get the result for the sum over all the numbers rather than just the odd ones? Well, here are two ways - one forwards and one backwards.

Now if we multiply the above by 1/4 we get:

Pi² /32 = 1/2² + 1/6² +...
Ie. the sum over all numbers with exactly one factor of 2.

Multiply by 1/4 again and we get:

Pi² /128 = 1/4² + 1/12² +...
Ie. the sum over all numbers with exactly two factors of 2.

Now repeat and sum them all up using the geometrical progression 1/4 + 1/16 + ... = 3/4. And we get:

Pi² /6 = 1/1² + 1/2² + 1/3² + ...

In the other direction. Suppose we assume that:

z = 1/1² + 1/2² + 1/3² + ...

Then z/4 = 1/2² + 1/4² + 1/6² + ...

Hence z-z/4 = 1/1² + 1/3² + 1/5² + ...

Thus 3z/4 = Pi² /8 (by the Fourier series stuff) and solving for z we get the answer we want.

AlexB.