How find this value$\sum\limits_{n=0}^{\infty}\frac{(2n)!}{(n!)^22^{3n+1}}$ show  that:
$$\sum_{n=0}^{\infty}\dfrac{(2n)!}{(n!)^22^{3n+1}}=\left(\frac{1}{2}\right)^{1/2}?$$
this sum is from  other problem,if I solve this,then the other problem is solve it
 A: Hint: what is $\frac{f^{(n)}(0)}{n!}$ when $f(x) = \sqrt{1-x}$?

Compute the first terms:
\begin{align}
f^{(0)}(x) &= \sqrt{1-x}\\
f^{(1)}(x) &= -\frac 12 (1-x)^{-1/2} \\
f^{(2)}(x) &= -\frac 14 (1-x)^{-3/2} \\
\end{align}
The exponent of $1-x$ in $f^{(n)}(x)$ is obviously $1/2 - n$. So when you derivate,
the front factor gets multiplied by $1/2 - n$ to get $f^{(n+1)}(x)$, and by $-1 $ because of the chain rule.
Hence the result is 
\begin{align}f^{(n+1)}(0) 
&= 1\times \left(\frac 12 - 0\right) \times \left(\frac 12 - 1\right)
\times \dots \times \left(\frac 12 - n\right) \times (-1)^{n+1} \\
&= \frac{1\times 3 \times \dots \times (2n-1)}{2^{n+1}} (-1)^n  \times (-1)^{n+1}
= -\frac{1\times 3 \times \dots \times (2n-1)}{2^{n+1}}\\
&= -\frac{(2n)!}{2^{n+1} 2\times 4 \times \dots \times 2n} = -\frac{(2n)!}{2^{2n+1} n!} \end{align}
You can check that this is true for the first values:
\begin{align}
f^{(1)}(x) &= -\frac 12 = 1\times \frac 12 &\times (-1)^1 \\
f^{(2)}(x) &= -\frac 14 = 1\times \frac 12 \times \left(-\frac 12\right)
&\times (-1)^2  
\end{align}
Then, you must show that the series is convergent for $x=\frac 12$ and you will get:
\begin{align}
\sqrt{1-x} 
&= 1 + \sum_{n=0}^\infty \frac{f^{(n+1)}(0) }{(n+1)!}x^{n+1} \\
&= 1 - \sum_{n=0}^\infty \frac{(2n)!}{2^{2n+1} (n!)^2 (n+1)} x^{n+1}
\end{align}

Unfortunately, when evaluated in $\frac 12$, this does not make appear the desired sum, because of the factor $n+1$.
We can get rid of it just using tern by term derivation:
\begin{align}
-\frac 12 (1-x)^{-1/2} 
&=- \sum_{n=0}^\infty \frac{(2n)!}{2^{2n+1} (n!)^2} x^n\\
 (1-x)^{-1/2} 
&= \sum_{n=0}^\infty \frac{(2n)!}{2^{2n} (n!)^2} x^n
\end{align}
And now we get, with $x=1/2$:
$$
 \left(1-\frac 12\right)^{-1/2} 
= \sum_{n=0}^\infty \frac{(2n)!}{2^{3n} (n!)^2} 
$$
A: Hint: Using the generalized binomial theorem, we get
$$
\begin{align}
\left(1-x\right)^{-1/2}
&=1+\frac12x+\frac{\frac12\cdot\frac32}{2!}x^2+\frac{\frac12\cdot\frac32\cdot\frac52}{3!}x^3+\frac{\frac12\cdot\frac32\cdot\frac52\cdot\frac72}{4!}x^4+\dots\\[6pt]
&=\sum_{k=0}^\infty\frac{(2k-1)!!}{2^kk!}x^k\\
&=\sum_{k=0}^\infty\frac{(2k)!}{2^{2k}k!^2}x^k\\
\end{align}
$$
Plug in $x=\frac12$ and check the result.
A: This is a binomial series. Try to rewrite $\bigg(\dfrac12\bigg)^\tfrac12$ as $\bigg(1-\dfrac12\bigg)^\tfrac12$ or as $\Big(1+1\Big)^{^{-\tfrac12}}$. Then expand, simplify the general term, and see what happens. :-$)$
