TITLE: Value of the binomial series $\sum_{i=0}^k \frac{{2i \choose i}}{4^i}$
QUESTION [6 upvotes]: 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!

REPLY [6 votes]: 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.