Evaluate the sum
x+23x3+23⋅45x5+23⋅45⋅67x7+…
Totally no idea. I think this series may related to the sinx series because of those missing even powers. Another way of writing this series:
∞∑k=0(2k)!!(2k+1)!!x2k+1.
Answer
In this answer, I mention this identity, which can be proven by repeated integration by parts:
∫π/20sin2k+1(x)dx=2k2k+12k−22k−1⋯23=12k+14k(2kk)
Your sum can be rewritten as
f(x)=∞∑k=01(2k+1)4k(2kk)x2k+1
Combining (1) and (2), we get
f(x)=∫π/20∞∑k=0sin2k+1(t)x2k+1dt=∫π/20xsin(t)dt1−x2sin2(t)=∫π/20−dxcos(t)1−x2+x2cos2(t)=−1√1−x2tan−1(xcos(t)√1−x2)]π/20=1√1−x2tan−1(x√1−x2)=sin−1(x)√1−x2
Radius of Convergence
This doesn't appear to be part of the question, but since some other answers have touched on it, I might as well add something regarding it.
A corollary of Cauchy's Integral Formula is that the radius of convergence of a complex analytic function is the distance from the center of the power series expansion to the nearest singularity. The nearest singularity of f(z) to z=0 is z=1. Thus, the radius of convergence is 1.
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