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### Infinite Series Involving Pi

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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!
```

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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/
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Associated Topics:
High School Sequences, Series

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