Hard Taylor Series identity
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Hard Taylor Series identity

[From: ] [author: ] [Date: 11-04-27] [Hit: ]
because even ns in the sum lead to sines of integer values of pi, which are 0; and for odd ns, 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. Ill leave that to you.......
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.

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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.
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keywords: Series,Hard,identity,Taylor,Hard Taylor Series identity
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