How can you prove that the sequence 1/n^2 converge to pi^2/6 ?
You probably mean that the sum of 1/n^2 converges to pi^2/6, the sequence itself converges to zero.
The problem itself is called Basel Problem and there are lots of methods to solve it but one of the easiest is the Plancherel identity which says the intergral of the square of the function is sum of the squares (of the absolute values) of the Fourier coefficients.
You do that for function x. (Namely f(x)=x)
then the integral from -pi to pi of x^2 is 2/3 pi^3
the n'th Fourier coefficient is 2pi/(i*n)
Cancelling the common factors gives you exactly the required identity
For more look in http://en.wikipedia.org/wiki/Basel_problem
Here is a very good answer to your question:
or : http://plus.maths.org/issue19/features/infseries/index.html
see page 5 in
or in : http://mathworld.wolfram.com/RiemannZetaFunctionZeta2.html
PI is a constant and N is a variable.
The converge point is:
The square over which you are integrating has sides with inclination
Pi/4 and -Pi/4. To integrate over this region, you should probably
split it into two regions as follows:
0 <= u <= sqrt(2)/2, -u <= v <= u,
sqrt(2)/2 <= u <= sqrt(2), u - sqrt(2) <= v <= sqrt(2) - u.
Now to do the inside integration with respect to v, write
2/(2+u^2-v^2) = A/(sqrt[2+u^2]+v) + B/(sqrt[2+u^2]-v),
A = -1/sqrt[2+u^2],
B = 1/sqrt[2+u^2].
Then the integral is
A*ln(sqrt[2+u^2]+v) + B*ln(sqrt[2+u^2]-v),
evaluated between the limits. This can be simplified to
Now the first term can be integrated, and you get
2*Arctanh[1/Sqrt(5)]^2, but the second term seems intractable.
A variation of the method you suggest is actually successful in
computing SUM 1/n^2. The trick is to compute the sum of the
reciprocals of the squares of the odd numbers only:
S = SUM 1/(2*n-1)^2
This can be shown to be identical to the double integral
S = INTEGRAL INTEGRAL 1/(1-x^2*y^2) dy dx,
using the same method. Now there is a very clever way to evaluate
this double integral: use the substitution
x = sin(u)/cos(v),
y = sin(v)/cos(u).
The Jacobian of this transformation is 1 - tan^2(u)*tan^2(v), and the
integrand becomes 1/(1-tan^2[u]*tan^2[v]), so the new integrand is
magically reduced to 1. The only problem is to figure out the new
limits of integration. It turns out that the region over which to
integrate is the triangle bounded by u = 0, v = 0, and u + v = Pi/2.
Integrating 1 over this region gives the area of the triangle, which
is obviously Pi^2/8 (half the base Pi/2 times the height Pi/2). Thus
S = Pi^2/8.
Now to find the actual sum you want, call it T, observe that
T - T/4 = S,
T = 4*S/3 = Pi^2/6.
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