Given the fact that sigma n=1 to infinity cos(nx)/n^2 = x^2/4 - pi*x/2 + pi^2/6 if 0 <= x <= 2pi , deduce the following.
sigma n=1 to infinity (-1)^(n+1) / (2n-1)^3 = pi^3 / 32
My work: I took the derivative of the Taylor series with respect to x which is justifiable to get sigma n=1 to infinity -nsin(nx) / n^2 = x/2 - pi/2. Then I let x = pi/2 so I could replace the sin(nx) with a (-1)^n, but I ended up with sigma (-1)^(n+1)/(2n-1) = pi/4 - pi/2 , which I can't seem to make similar to the desired identity.
sigma n=1 to infinity (-1)^(n+1) / (2n-1)^3 = pi^3 / 32
My work: I took the derivative of the Taylor series with respect to x which is justifiable to get sigma n=1 to infinity -nsin(nx) / n^2 = x/2 - pi/2. Then I let x = pi/2 so I could replace the sin(nx) with a (-1)^n, but I ended up with sigma (-1)^(n+1)/(2n-1) = pi/4 - pi/2 , which I can't seem to make similar to the desired identity.
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Integrate from x=0 to x=t, for 0< t < 2pi.
If you can justify inverting the Sum and the Integral, then you'd get:
Sum(n=1 to inf; sin(nt) /n^3 ) = (1/12) t^3 - (1/4) pi t^2 + (1/6) pi^2 t
and evaluating at t=pi/2:
Sum(n=1 to inf; sin(n pi/2) /n^3) = pi^3 /32
or
Sum(n=1 to inf; (-1)^(n+1) /(2n-1)^3 ) = pi^3 /32
because even n's in the sum lead to sines of integer values of pi, which are 0; and for odd n's, the sign alternates, whether we are at pi/2 or 3pi/2 modulo 2pi.
All that is left is to justify switching the integration and the sum. I'll leave that to you.
If you can justify inverting the Sum and the Integral, then you'd get:
Sum(n=1 to inf; sin(nt) /n^3 ) = (1/12) t^3 - (1/4) pi t^2 + (1/6) pi^2 t
and evaluating at t=pi/2:
Sum(n=1 to inf; sin(n pi/2) /n^3) = pi^3 /32
or
Sum(n=1 to inf; (-1)^(n+1) /(2n-1)^3 ) = pi^3 /32
because even n's in the sum lead to sines of integer values of pi, which are 0; and for odd n's, the sign alternates, whether we are at pi/2 or 3pi/2 modulo 2pi.
All that is left is to justify switching the integration and the sum. I'll leave that to you.