How to show $\sum_{i=1}^{\lceil L/2\rceil}\frac{\binom{2i-2}{i-1}}{i}2^{-2i}=1/2-2^{-L-1}\binom{L}{(L-1)/2}$? By numerical evaluation, it seems that the following identity holds. For any odd positive integer $L$,
$$\sum_{i=1}^{\lceil L/2\rceil}\frac{\binom{2i-2}{i-1}}{i}2^{-2i}=1/2-2^{-L-1}\binom{L}{(L-1)/2}.$$
I was trying to expand everything out but didn't have a good luck. I guess there should be a more clever way using some binomial identity. 
 A: Building on my comment, this identity can be proven by induction. (Though I feel like there is some interesting combinatorial argument here, given that the Catalan numbers are appearing in the sum.)
So we will prove the identity
$$\sum_{i=0}^{K}\frac{\binom{2i}{i}}{i+1}2^{2(K-i)}=2^{2K+1}-\binom{2K+1}{K}$$
by induction on $K$. For $K=0$, this is trivial as both sides are 1.
Now suppose the statement holds for some $K\geq 0$. Then 
$$\begin{align*}
\sum_{i=0}^{K+1}\frac{\binom{2i}{i}}{i+1}2^{2(K+1-i)}
  &\stackrel{(a)}{=}2^2\sum_{i=0}^{K}\frac{\binom{2i}{i}}{i+1}2^{2(K-i)}+\frac{1}{K+2}\binom{2K+2}{K+1}
\\&\stackrel{(b)}{=}2^2\left(2^{2K+1}-\binom{2K+1}{K}\right)+\frac{1}{K+2}\binom{2K+2}{K+1}
\\&\stackrel{(c)}{=}2^{2(K+1)+1}-4\binom{2K+1}{K}+\frac{1}{K+2}\binom{2K+2}{K+1}
\\&\stackrel{(d)}{=}2^{2(K+1)+1}-2\left(\binom{2K+1}{K}+\binom{2K+1}{K+1}\right)+\frac{1}{K+2}\binom{2K+2}{K+1}
\\&\stackrel{(e)}{=}2^{2(K+1)+1}-\frac{2K+3}{K+2}\binom{2K+2}{K+1}
\\&\stackrel{(f)}{=}2^{2(K+1)+1}-\binom{2(K+1)+1}{(K+1)+1}
\end{align*}$$
where (a) follows from peeling off the last summand, (b) follows from pulling out a power of two and applying induction, (c) follows from distributing the power of 2, (d) follows from the identity $\binom{n}{k}=\binom{n}{n-k}$, (e) follows from the identity $\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}$, and (f) follows from the factorial definition of $\binom{n}{k}$.
A: When looking at OPs identity the  left-hand side smells like  a telescoping sum where we have something like the first and last entry at the right hand side, and indeed we can transform it accordingly.  We start with some preliminary arrangements.  Shifting the index $i$ by one to start with $i=0$ we want to show
\begin{align*}
\sum_{i=0}^{\lceil L/2\rceil -1}\frac{1}{i+1}\binom{2i}{i}2^{-2i-2}=\frac{1}{2}-\frac{1}{2^{L+1}}\binom{L}{\frac{L-1}{2}}\tag{1}
\end{align*}
Since $L$ is assumed to be an odd positive integer, we set $L=2K+1$ in (1), multiply both sides with $4$ and show

The following is valid for non-negative integer $K$:
  \begin{align*}
\sum_{i=0}^K\frac{1}{4^i(i+1)}\binom{2i}{i}=2-\frac{1}{4^K}\binom{2K+1}{K}\tag{2}
\end{align*}
We obtain
  \begin{align*}
\color{blue}{\sum_{i=0}^K}&\color{blue}{\frac{1}{4^i(i+1)}\binom{2i}{i}}\\
&=1+\sum_{i=1}^K\frac{1}{4^i}\left(\binom{2i}{i}-\binom{2i}{i-1}\right)\tag{3}\\
&=1+\sum_{i=1}^K\frac{1}{4^i}\left(4\binom{2i-1}{i-1}-\binom{2i+1}{i}\right)\tag{4}\\
&=1+\sum_{i=1}^K\frac{1}{4^{i-1}}\binom{2i-1}{i-1}-\sum_{i=1}^K\frac{1}{4^i}\binom{2i+1}{i}\tag{5}\\
&=1+\sum_{i=0}^{K-1}\frac{1}{4^{i}}\binom{2i+1}{i}-\sum_{i=1}^K\frac{1}{4^i}\binom{2i+1}{i}\tag{6}\\
&=1+1-\frac{1}{4^K}\binom{2K+1}{K}\tag{7}\\
&\,\,\color{blue}{=2-\frac{1}{4^K}\binom{2K+1}{k}}
\end{align*}
and OPs claim follows.

Comment:


*

*In (3)   we  use the binomial identity $\frac{1}{i+1}\binom{2i}{i}=\binom{2i}{i}-\binom{2i}{i-1}$  which might be familiar in context with the Catalan numbers.

*In (4) we use the identity
\begin{align*}
\binom{2i}{i}-\binom{2i}{i-1}&=2\binom{2i}{i}-\binom{2i}{i}-\binom{2i}{i-1}\\
&=2\binom{2i}{i}-\binom{2i+1}{i}\\
&=4\binom{2i-1}{i-1}-\binom{2i+1}{i}
\end{align*}

*In (5)  we split the sum.

*In (6) we shift the index  of the  left-hand sum by one  to start with $i=0$.

*In (7)  we cancel  terms due to the telescoping sum.
