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.
On
Seeking to show that (problem statement by @SeanClark)
$$\sum_{q=0}^K \frac{1}{q+1} {2q\choose q} 2^{2K-2q} = 2^{2K+1} - {2K+1\choose K}$$
We get the difference of two pieces from the Catalan number, namely
$$\sum_{q=0}^K {2q\choose q} 2^{2K-2q} \quad\text{and}\quad \sum_{q=0}^K {2q\choose q+1} 2^{2K-2q}.$$
These are
$$\sum_{q=0}^K {2K-2q\choose K-q} 2^{2q} \quad\text{and}\quad \sum_{q=0}^K {2K-2q\choose K+1-q} 2^{2q}.$$
The first one is
$$\sum_{q=0}^K 2^{2q} [z^{K-q}] (1+z)^{2K-2q} = [z^K] (1+z)^{2K} \sum_{q=0}^K 2^{2q} z^q (1+z)^{-2q}.$$
Now we may extend $q$ beyond $K$ due to the coefficient extractor in front:
$$[z^K] (1+z)^{2K} \sum_{q\ge 0} 2^{2q} z^q (1+z)^{-2q} = [z^K] (1+z)^{2K} \frac{1}{1-2^2z/(1+z)^2} \\ = [z^K] (1+z)^{2K+2} \frac{1}{(1+z)^2-2^2z} = [z^K] (1+z)^{2K+2} \frac{1}{(1-z)^2}.$$
The second one is
$$-{-2\choose 0} 2^{2K+2} + \sum_{q=0}^{K+1} {2K-2q\choose K+1-q} 2^{2q} = - 2^{2K+2} + \sum_{q=0}^{K+1} 2^{2q} [z^{K+1-q}] (1+z)^{2K-2q} \\ = - 2^{2K+2} + [z^{K+1}] (1+z)^{2K} \sum_{q=0}^{K+1} 2^{2q} z^q (1+z)^{-2q} \\ = - 2^{2K+2} + [z^{K+1}] (1+z)^{2K+2} \frac{1}{(1-z)^2}.$$
Now subtracting we have
$$[z^K] (1+z)^{2K+2} \frac{1}{(1-z)^2} + 2^{2K+2} - [z^{K+1}] (1+z)^{2K+2} \frac{1}{(1-z)^2} \\ = [z^{K+1}] (1+z)^{2K+2} \frac{z}{(1-z)^2} + 2^{2K+2} - [z^{K+1}] (1+z)^{2K+2} \frac{1}{(1-z)^2} \\ = 2^{2K+2} + [z^{K+1}] (1+z)^{2K+2} \frac{z-1}{(1-z)^2} = 2^{2K+2} - [z^{K+1}] (1+z)^{2K+2} \frac{1}{1-z}.$$
The remaining coefficient is not difficult to do and we find
$$2^{2K+2} - \sum_{q=0}^{K+1} {2K+2\choose q} = 2^{2K+2} - {2K+2\choose K+1} - \sum_{q=0}^{K} {2K+2\choose q} \\ = 2^{2K+2} - {2K+2\choose K+1} - \frac{1}{2} \left(2^{2K+2} - {2K+2\choose K+1} \right) = 2^{2K+1} - \frac{1}{2} {2K+2\choose K+1} \\ = 2^{2K+1} - \frac{1}{2} \frac{2K+2}{K+1} {2K+1\choose K}.$$
This is indeed
$$\bbox[5px,border:2px solid #00A000]{ 2^{2K+1} - {2K+1\choose K}.}$$
as claimed.
On
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.
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 $\binom{n}{k}$.