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I am trying to prove that the Taylor expansion of $\arcsin(x) = \sum\limits_{n=0}^\infty \cfrac{(2n!)x^{2n+1}}{(2^nn!)^2(2n+1)}$. Sorry about the notation, I'm not sure what syntax to use. S stands for sigma.

My understanding is that letting $f(x) = \arcsin(x)$, $f'(x) = \cfrac{1}{\sqrt{1-x^2}}$

so using the binomial series for $(1+x)^k = 1 + kx + \cfrac{1}{2!}k(k-1)x^2 + ...$

I got $(1+(-x^2))^{-1/2}$ and tried to substitute with $k = -1/2$ and $x$ with $(-x^2)$.

Once I get the formula I integrate $f'(x)$, solve for the constant value when $x = 0$ and done.

I keep getting a different solution and I'm afraid I'm doing this wrong. Can someone help me out ?

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For some basic information about writing math at this site see e.g. here, here, here and here. – Américo Tavares Apr 6 '13 at 0:24
What did you get? Are you aware that $(2n)!/(2^n\,n!)=(2n-1)(2n-3)...5\cdot3\cdot1$? – Berci Apr 6 '13 at 0:26
Wow, that blew my mind ! It helped a lot and I was able to prove it. Thanks ! – hyg17 Apr 6 '13 at 21:05

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