Value of the binomial series $\sum_{i=0}^k \frac{{2i \choose i}}{4^i}$ Some time ago a question was asked here regarding the value of the sum $$\sum_{i=0}^k \frac{{2i \choose i}}{4^i}$$. 
But it was deleted later by the OP. I went around it but didn't find a solution.
Some common combinatoric identities I know of are $4^i=2^{2i}=\sum_{j=0}^{2i} {2i\choose j}$ also that ${2i \choose i}=\sum_{j=0}^{i} {i\choose j}^2$. But they were hardly of any use. I also thought of individual term as probability but turns out I can't think of it as anything sensible.a quick WA check gave the answer as $${k+\frac{1}{2} \choose k}=\frac{(2k+1)!}{k!^24^k}$$
Help would be appreciated!
 A: In another way, using the duplication formula for Gamma
$$
\Gamma \left( {2\,z} \right) = {{2^{\,2\,z - 1} } \over {\sqrt \pi  }}\Gamma \left( z \right)\Gamma \left( {z + 1/2} \right)
 = 2^{\,2\,z - 1} {{\Gamma \left( z \right)\Gamma \left( {z + 1/2} \right)} \over {\Gamma \left( {1/2} \right)}}
$$
we have
$$
{1 \over {4^{\,i} }}\binom{2i}{i}
 = {1 \over {4^{\,i} }}{{\Gamma \left( {2\left( {i + 1/2} \right)} \right)} \over {\Gamma \left( {i + 1} \right)^{\,2} }}
 = {{\Gamma \left( {i + 1/2} \right)} \over {\Gamma \left( {1/2} \right)\Gamma \left( {i + 1} \right)}} = \binom{i-1/2}{i}
$$
Then our sum is
$$
\eqalign{
  & \sum\limits_{i = 0}^k {{1 \over {4^{\,i} }}\left( \matrix{
  2i \cr 
  i \cr}  \right)}  =   \cr 
  &  = \sum\limits_{i = 0}^k {\left( \matrix{
  i - 1/2 \cr 
  i \cr}  \right)}  = \quad \quad (1)  \cr 
  &  = \sum\limits_i {\left( \matrix{
  k - i \cr 
  k - i \cr}  \right)\left( \matrix{
  i - 1/2 \cr 
  i \cr}  \right)}  = \quad \quad (2)  \cr 
  &  = \left( \matrix{
  k + 1/2 \cr 
  k \cr}  \right)\quad \quad (3) \cr} 
$$
where:
 1) for the identity above;
 2) replacing the sum upper bound with the first b., the lower bound is implicit in the second b.;
 3) using the "double convolution" formula for binomials.  
A: We prove that $$\sum_{k=0}^{n} \frac{{2k \choose k}}{4^k} = \frac{(2n+1)!}{n!^24^n}.$$
We proceed by induction. The base case is easily verifiable. Denote the RHS of the above equation as $a_n$.
Now consider $$a_{n+1} = \frac{(2n+3)!}{(n+1)!^24^{n+1}} = \frac{(2n+2)(2n+3)}{4(n+1)^2} \cdot \frac{(2n+1)!}{n!^24^n} = \frac{(2n+3)}{2n+2} a_n.$$
So we now have $$a_{n+1} - a_n =\frac{a_n}{2n+2} = \frac{(2n+1)!}{2(n+1)\cdot n!^24^n} = \frac{(2n+1)!}{(n+1)!n!4^{n+1}}\cdot 2\\ = \frac{(2n+1)!}{(n+1)!n!4^{n+1}}\cdot \frac{2n+2}{n+1} = \frac{(2n+2)!}{(n+1)!^24^{n+1}} = \frac{{2(n+1) \choose n+1}}{4^{n+1}},$$
which is the added term in the summation. This completes the induction.
A: We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. Recalling the generating function of the central binomial coefficients we can write 
\begin{align*}
[z^i]\frac{1}{\sqrt{1-4z}}=\binom{2i}{i}\tag{1}
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{i=0}^{k}}\color{blue}{\binom{2i}{i}\frac{1}{4^i}}
&=\sum_{i=0}^{k}[z^i]\frac{1}{\sqrt{1-z}}\tag{2}\\
&=[z^0]\frac{1}{\sqrt{1-z}}\sum_{i=0}^{k}z^{-i}\tag{3}\\
&=[z^0]\frac{1}{\sqrt{1-z}}\cdot\frac{\left(\frac{1}{z}\right)^{k+1}-1}{\frac{1}{z}-1}\tag{4}\\
&=[z^k]\frac{1-z^{k+1}}{(1-z)^{3/2}}\tag{5}\\
&=[z^k]\sum_{j=0}^\infty\binom{-\frac{3}{2}}{j}(-z)^j\tag{6}\\
&=[z^k]\sum_{j=0}^\infty\binom{j+\frac{1}{2}}{j}z^j\tag{7}\\
&\,\,\color{blue}{=\binom{k+\frac{1}{2}}{k}}
\end{align*}
and the claim follows.

Comment:


*

*In (2) we apply the coefficient of operator according to (1).

*In (3) we use the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

*In (4) we apply the finite geometric series formula.

*In (5) we do some simplifications and apply the rule from (3) again.

*In (6)  we observe $z^{k+1}$ in the numerator does not contribute to $[z^k]$ and we make a binomial series expansion.

*In (7) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

*In (8) we finally select the coefficient of $z^k$.
