Let F(x)=Sum(n=1-->infinity)[ {x^(n-1)}/{n^(2)*(1+x^(n))} ]
find Integral (x=0-->1)[F(x)dx]?
A question
Re: A question from Gre Sub exam
I got very strange answer:
= int(sum) = sum(int) = sum(int(d(1+x^n)/{n^3*(1+x^n)})) = sum((1/n)*log(1+x^n)|{from 0 to 1}) =
= log2*sum(1/n^3) = log2 * zeta(3),
where zeta(3) - Apéry's_constant
http://en.wikipedia.org/wiki/Apéry's_constant
= int(sum) = sum(int) = sum(int(d(1+x^n)/{n^3*(1+x^n)})) = sum((1/n)*log(1+x^n)|{from 0 to 1}) =
= log2*sum(1/n^3) = log2 * zeta(3),
where zeta(3) - Apéry's_constant
http://en.wikipedia.org/wiki/Apéry's_constant
Re: A question from Gre Sub exam
i would think that the answer would not be finite, since the integral of each of individual terms is infinite. however, i am not sure about being able to exchange sum and the integral here though.
> int(x^(n-1)/(1-x^n), x = 0 .. 1);
infinity
---------
signum(n)
> int(x^(n-1)/(1-x^n), x = 0 .. 1);
infinity
---------
signum(n)
Re: A question from Gre Sub exam
Well, |x^(n-1)/(1 + x^n)| < 1 for all x on [0, 1], so |1/n^2 * x^(n-1)/(1 + x^n)| < 1/n^2 on [0, 1]. Since 1/1^2 + 1/2^2 + 1/3^2 + ... converges to zeta(2), the series x^n/[n^2(1 + x^n)] is uniformly convergent by the Weierstrass M-test and so the sum and limit operators should commute.
Re: A question from Gre Sub exam
Forget about it. Original problem sounded a little bit different and had more decent solution. But we will not discuss it here to prevent breaking the ETS policy.