<s>[INST] 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. 
 [/INST] 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.
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