Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Infinite Series Involving Pi


Date: 12/16/97 at 01:00:59
From: Eric Tak
Subject: Infinite series involving pi

Sorry, I am just totally confused on this problem.  Here it is:

   1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 + . . . . . = pi^2/6

Please, I need a clear, easy-to-understand reason or proof for WHY
this is correct. If a simple solution is unavailable, a hard solution
will also be greatly appreciated. 

Also, what does:

   1 - 1/2^2 + 1/3^2 - 1/4^2 + 1/5^2 - . . .

equal?

Thank you!


Date: 12/16/97 at 19:30:24
From: Doctor Anthony
Subject: Re: Infinite series involving pi

This can be proved by advanced trig. manipulations, but a more elegant 
method is with the aid of Fourier series.

A brief note on Fourier series is necessary if you are to understand 
the method.

For a great many functions within the range  -pi < x < pi we can 
express f(x) as an infinite series of the form:

 f(x) = a(0)/2 + SUM(n=1 to inf.)[a(n)cos(nx) + b(n)sin(nx)]

where the a(i), b(i) are constants.  So

 f(x) = a(0)/2 + a(1)cos(x) + a(2)cos(2x) + a(3)cos(3x) + ....
               + b(1)sin(x) + b(2)sin(2x) + b(3)sin(3x) + ....


This series is periodic with period 2.pi and the convention is to take 
the interval -pi < x < pi.

The way to evaluate the constants a(i),is to multiply both sides by 
cos(nx) and then integrate between -pi and pi.

  INT(-pi to pi)[f(x)cos(nx)dx] = pi.a(n)   (proof given below)

   So a(n) = (1/pi)INT(-pi to pi)[f(x)cos(nx)dx]

Similarly

      b(n) = (1/pi)INT(-pi to pi)[f(x)sin(nx)dx]

and   a(0) = (1/pi)INT(-pi to pi)[f(x).dx]


Proof of above results.

INT(-pi to pi)[cos(mx)cos(nx)dx]   m not equal n

   = (1/2)INT[cos(m+n)x + cos(m-n)x]dx  between -pi and pi 

   = 0    since sin(any multiple of pi) = 0

but  INT(-pi to pi)[cos^2(nx)]dx = pi

So all the terms cos(mx)cos(nx)  (m not equal to n) disappear and we 
get only the term in cos^2(nx)

This gives the result  INT[f(x)cos(nx)dx] = a(n).pi

and so a(n) = (1/pi)INT[f(x).cos(nx).dx]

In the case n=0 we get   INT[f(x).dx] = INT[a(0)/2)dx] + 0 + 0+ ..

                                       = a(0).pi

So     a(0) = (1/pi)INT[f(x).dx]

Similarly  b(n) = (1/pi)INT[f(x)sin(nx)dx]

since all terms in sin(mx)sin(nx) (m not equal to n) disappear leaving 
only the term in sin^2(nx).  Also all terms in sin(mx)cos(nx) whether 
or not m=n disappear on integration between -pi and pi.

This means that we can evaluate all the a(i), b(i) and express f(x) as 
an infinite series in sines and cosines of multiples of x. There are a 
great many functions that can be expressed in this way in the interval 
-pi to pi.

We can now turn our attention to finding the sum of the series 
1 + 1/2^2 + 1/3^2 + 1/4^2 + ....

Let f(x) = x^2   This is an even function of x so we know that NONE of 
the sine terms of the Fourier series can be present. We need only look 
at the a(i) terms.

For all a(n)  we get pi.a(n) = INT(-pi to pi)[x^2.cos(nx)dx]

When  n = 0 this gives    a(0) = (2/3)pi^2

When n greater than 0 integration by parts in 3 steps gives

                      a(n) = 4(-1)^n/n^2

The Fourier series is  pi^2/3 + 4.SUM(1 to infinity)(-1)^n/n^2 cos(nx)

With x=pi, we get

      pi^2 = pi^2/3 + 4.SUM(-1)^n/n^2 cos(n.pi)

and since cos(n.pi)= (-1)^n  this produces +'s for each term

      2.pi^2/3  =  4.SUM[1 + 1/2^2 + 1/3^2 + ..... to infinity]

         pi^2/6 =  1 + 1/2^2 + 1/3^2 + 1/4^2 + ..... to infinity. 

As for the other series, first note that:

      1 - 1/4 + 1/9 - 1/16 + 1/25 - ....... = pi^2/12

Then let f(x) = x^2. This is an even function of x so we know that NONE of 
the sine terms of the Fourier series can be present. We need only look at 
the a(i) terms.

For all a(n) we get pi.a(n) = INT(-pi to pi)[x^2.cos(nx)dx]

When  n = 0 this gives  a(0) = (2/3)pi^2

When n greater than 0 integration by parts in 3 steps gives:

      a(n) = 4(-1)^n/n^2

The Fourier series is  pi^2/3 + 4.SUM(1 to infinity)(-1)^n/n^2 cos(nx)

With x=0, we get:

      0 = pi^2/3 + 4.SUM(-1)^n/n^2 cos(0)

and since cos(0) = 1 this produces alternate + and - for the terms.

      -pi^2/3  =  4.SUM[-1 + 1/2^2 - 1/3^2 + ..... to infinity]

      -pi^2/12 =  -1 + 1/2^2 - 1/3^2 + 1/4^2 + ..... to infinity. 

       pi^2/12 =  1 - 1/4 + 1/9 - 1/16 + ...  to infinity

(Well you did ask for a 'hard' solution.)

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Sequences, Series

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/