I can't solve this series I was given the following series
$\
S(n) = 1+\frac{1}{2}\cdot \frac{1}{2}+
\frac{3*1}{4*2}\cdot\left ( \frac{1}{2}\right )^2 + \frac{5*3*1}{6*4*2}\left ( \frac{1}{2} \right )^3 + ... $
i noticed that each individual term looks very similar to the solution for a wallis integral
Thus
$$
S(n) = \frac{2}{\pi}\int_{0}^{\pi/2}\frac{sin^0x}{2^0} + \frac{sin^2x}{2^1}+\frac{sin^4x}{2^2}+\frac{sin^6x}{2^3}+...dx
$$
Which appeared to be a geometric series,
Hence
$$ 
S(n) = \frac{2}{\pi}\int_{0}^{\pi/2} 
\frac{(\frac{sin^2x}{2})^n-1}{\frac{sin^2x}{2}-1}
dx
$$
Which after some algebraic simplifications gives me this horrible integral at which i am stuck
$$
= \frac{2^{2 - n}}{\pi}\int_{0}^{\pi/2}\frac{sin^{2n}x - 2^n }{sin^2x - 2} dx
$$
So how can this integral be solved ? is it non-elementary integral ? Or is there another approach to solve this problem ?
Edit:
Thanks for your response, i thought at first that the question wanted me to find the sum of the series to the n-th term,
In continuing to my previous approach i would be using infinite geometric series formula instead
$$
S =\frac{2}{π} \int_{0}^{\pi/2} \frac{dx}{1 - \frac{sin^2x}{2}} = √2
$$
Which can be easily solved by different techniques !
 A: Examine the coefficients more closely:
$$\frac{5\cdot3\cdot1}{6\cdot4\cdot2}=\frac{6!}{(6\cdot4\cdot2)^2}=\frac{6!}{(2^3\cdot 3!)^2}=\frac{6!}{(3!)(3!)}\cdot\frac{1}{2^{2\cdot 3}}=\binom{6}{3}\frac{1}{2^{2\cdot3}}$$
If the pattern is consistent as you suggest, we have the $n$th coefficient (considered as a power series in $1/2$):
$$\binom{2n}{n}\frac{1}{2^{2n}}$$
Which you may recognise (from the derivation of the $\arcsin$ series) as the binomial series of the square root function: $$(1-x)^{-1/2}$$
So, this is just $(1-1/2)^{-1/2}=\sqrt{2}$.
N.B: Why is that the series of $(1-x)^{-1/2}$? The derivatives of this function $f(x)=(1-x)^{-1/2}$ at $x=0$ are of the form (don't forget the chain rule with the $-x$ in the middle!):
$$\begin{align}f(0)&=1^{-1/2}=1\\f'(0)&=-\frac{1}{2}(-1)\cdot1^{-3/2}=\frac{1}{2}\\f''(0)&=-\frac{1}{2}(-1)\cdot-\frac{3}{2}(-1)\cdot1^{-5/2}=\frac{1\cdot3}{2\cdot2}\\&\cdots\\f^{(n)}(x)&=\frac{(2n-1)!!}{2^n}(1-x)^{-(2n+1)/2},\,f^{(n)}(0)=\frac{(2n-1)!!}{2^n}\end{align}$$
Try showing this by induction, and then using $f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}\cdot x^n$ try showing the coefficients match the form I gave before.
