Convergence of Euler's Infinite Series

Date: 03/06/2002 at 21:00:17
From: Chris 
Subject: The Convergence of one of Euler's Infinite Series

I was writing to inquire about the convergence of the infinite series
1+(1/2^2)+(1/3^2)+(1/4^2)+(1/5^2)+(1/6^2)+...+(1/n^2) = ?

The hint given along with this problem was that the answer has 
something to do with the quadrature of a circle. I am a little bit 
stumped by this comment but believe it has something to do with pi.

My first attempt at solving this problem was to try to split it 
between the even and odd denominators. So 

   [(1/2^2)+(1/4^2)+(1/6^2)+...] + [1+(1/3^2)+(1/5^2)+(1/7^2)...]

From here i attempted to find the sum of each part, but both have 
escaped me.

My second attempt was to approximate the decimal value and work 
backward. Using a computer program I approximated the decimal value 
to 1.6449, but this is not a recognizable fraction and it does not 
seem to be growing fast enough to approach 1.666666666

Any help in solving this series would be greatly appreciated, 
preferably an answer and perhaps some general advice when solving 
infinite series.  

Thank you in advance for your time!
Chris

Date: 03/07/2002 at 00:52:17
From: Doctor Schwa
Subject: Re: The Convergence of one of Euler's Infinite Series

Hi Chris,

The only solutions I know to this sum uses quite a bit of calculus.
One nice method is in our archives at

  Sum of 1/n^2
  http://mathforum.org/dr.math/problems/schmidt.7.24.00.html   

and my favorite method is at

  Euler's summmation of 1/n^2
  http://mathforum.org/dr.math/problems/bennett.2.15.00.html   

and yet another method is at

  Infinite Series Involving Pi
  http://mathforum.org/dr.math/problems/tak12.16.97.html   

Give those a try, and feel free to write back if you still have 
questions.

- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/   

Date: 03/07/2002 at 19:10:31
From: Chris 
Subject: Re: The Convergence of one of Euler's Infinite Series

Dear Dr. Schwa,

Thank you many many times for your help with that problem. Sadly, I 
don't fully grasp the solutions you presented and I am afraid the 
problem is on my end. To aid in my understanding, do you think you 
could explain the "infinite polynomial product" theorem and the "power 
series expansion" you used in your solution of the problem (the second 
link wherein you talked about the function f(x) = sin (pix)? Neither 
my previous math courses nor my current calculus class have introduced 
these concepts yet.

Thank you in advance once again!
Chris

Date: 03/07/2002 at 19:31:25
From: Doctor Schwa
Subject: Re: The Convergence of one of Euler's Infinite Series

Hi Chris,

The infinite polynomial part is within your reach from only an
algebra class.

sin(pi x) is zero for all integers x. Thus, just as a polynomial with 
k as a zero has to have (1 - x/k) as a factor, so must this function.

Perhaps you're more used to seeing it as (x-k), but in this case that 
would create a function that blows up to infinity. Dividing through 
by k maintains the zero and gives a finite function that you can then 
set to the right height by matching the slope at x = 0 (which is pi).

The power series expansion comes toward the end of calculus BC, but 
the idea is a generalization of the linear approximation. That is,

sin (x) is approximately sin(0) + x * (the derivative of sin at 0),
or more generally,
f(x) = about f(0) + x * f'(0).
That just means to follow the tangent line.

You can generalize this, to follow the parabola, or best cubic,
or so on, and find 

f(x) = approximately 
f(0) + x * f'(0) + (x^2 / 2) f'''(0) + (x^3 / 6) f'''(0) + ...

from which you get

sin(pi x) = approximately 0 + pi x * 1 + 0 - (x^3 / 6) pi^3 + ...

Then you can set the coefficients of x, and of x^3, equal to each
other in these two different methods ...

Does that help clear things up?

- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/   

Date: 03/08/2002 at 18:18:58
From: Chris 
Subject: Re: The Convergence of one of Euler's Infinite Series

Dear Dr. Schwa, 

Thank you very much for the clarification; that makes the solution 
infinitely (no pun intended) clearer. That pretty much answers my 
questions. Thank you very much for your time; I think you and all the 
Dr. Math people are doing a great service by helping to stimulate the 
desire to do mathematics.

Thanks again!
Chris