Questions We consider the series sumn=1^infty...

Questions We consider the series $$sum_{n=1}^{infty} frac{1}{n^{2}}$$ (2). Prove that $$sum_{k=0}^{infty} frac{1}{(2 k+1)^{2}}=int_{0}^{1} int_{0}^{1} frac{1}{1-x^{2} y^{2}} d x d y$$

Transcript

00:01 So in this problem, we're given, we're told to consider the series, the sum over 1 over n squared.

00:08 But this is already a later part of the problem.

00:12 And we're actually asked to prove that the sum over k equals 0 to infinity of 1 over 2k plus 1 squared is equal to the integral from 0 to 1 of the integral from 0 to 1.

00:31 Of the function 1 over 1 minus x squared y squared dy dx.

00:39 And so the way we're going to do this is we're going to look at the integrand, and we're going to expand that as a sum, and then we can swap the sum and the integral around using the fact that this integral is a linear function, and calculate the integral that way.

00:58 So let's consider the integrand 1 1 over 1 minus x squared y squared.

01:05 So if you know about your infinite series, you may recognize this as being related to the geometric series.

01:13 So this is a well -known fact that if you take the sum from n equals 0 to infinity of r to the n, where this is equal to 1 over 1 minus r as long as r is less the modulus of r is less than or equal to 1 because this integrand is happening over the range 0 to 1 for both x and y we know x and y falls within this range and so we can expand this using a geometric series as the sum over k k equals zero to infinity of x squared y squared to the n, which we can simply rewrite as the sum over k equals zero to infinity of x to the 2n times y to the 2n.

02:09 Now, if we take the double integral, we have the double integral from zero to one, the integral from 0 to 1 of the sum over k equals 0 to infinity x to the 2n y to the 2n dx dy.

02:30 Now we can pull the sum out of the front because again the integral is a linear function and so we're left with the sum from k equals 0 to infinity and because the function is actually separable we can write this as just the integral from 0 to 1 of x to the 2n dx times the integral from 0 to 1 of y to the 2n dy and now this integral is straightforward to perform so we know this is now equal to the sum from k k equals zero to infinity of x to the 2n plus 1 over 2n plus 1, evaluated between 1 and 0.

03:18 And again, now multiply by y to the 2n plus 1 over 2n plus 1, evaluated between 1 and 0...

Question

Questions We consider the series $$sum_{n=1}^{infty} frac{1}{n^{2}}$$ (2). Prove that $$sum_{k=0}^{infty} frac{1}{(2 k+1)^{2}}=int_{0}^{1} int_{0}^{1} frac{1}{1-x^{2} y^{2}} d x d y$$

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