# Proving 1/n^2 converge to pi^2/6

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

"Nothing worth knowing can be taught" Oscar Wilde

Order by

PI is a constant and N is a variable.

The converge point is:

1/n^2=pi^2/6

1/n=(pi^2/6)^(1/2)

n= 1/((pi^2/6)^(1/2))

The power of Simplicity

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, and 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 4*Arctanh(u/sqrt[2+u^2])/sqrt(2+u^2) + 4*Arctanh([sqrt(2)-u]/sqrt[2+u^2])/sqrt(2+u^2). 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: infinity S = SUM 1/(2*n-1)^2 n=1 This can be shown to be identical to the double integral 1 1 S = INTEGRAL INTEGRAL 1/(1-x^2*y^2) dy dx, 0 0 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.

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, and 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 4*Arctanh(u/sqrt[2+u^2])/sqrt(2+u^2) + 4*Arctanh([sqrt(2)-u]/sqrt[2+u^2])/sqrt(2+u^2). 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: infinity S = SUM 1/(2*n-1)^2 n=1 This can be shown to be identical to the double integral 1 1 S = INTEGRAL INTEGRAL 1/(1-x^2*y^2) dy dx, 0 0 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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