When working on a problem I was faced with the following binomial identity valid for integers $m,n\geq 0$:
\begin{align*}
\color{blue}{\sum_{l=0}^m(-4)^l\binom{m}{l}\binom{2l}{l}^{-1}
\sum_{k=0}^n\frac{(-4)^k}{2k+1}\binom{n}{k}\binom{2k}{k}^{-1}\binom{k+l}{l}
=\frac{1}{2n+1-2m}}\tag{1}
\end{align*}I have troubles to prove it and so I'm kindly asking for support.
Maybe the following simpler one-dimensional identity could be useful for a proof. We have for non-negative integers $n$:
\begin{align*}
\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2k+1}=\frac{4^{n}}{2n+1}\binom{2n}{n}^{-1}\tag{2}
\end{align*}
The LHS of (2) can be transformed to
\begin{align*}
\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2k+1}&=\sum_{k=0}^n(-1)^k\binom{n}{k}\int_{0}^1x^{2k}dx\\
&=\int_{0}^1\sum_{k=0}^n(-1)^k\binom{n}{k}x^{2k}\,dx\\
&=\int_{0}^1(1-x^2)^n\,dx
\end{align*}
Using a well-known integral representation of reciprocals of binomial coefficients the RHS of (2) can be written as
\begin{align*}
\frac{4^{n}}{2n+1}\binom{2n}{n}^{-1}&=4^n\int_{0}^1x^n(1-x)^n\,dx
\end{align*}
and the equality of both integrals can be shown easily. From (2) we can derive a simple one-dimensional variant of (1).
We consider binomial inverse pairs and with respect to (2) we obtain
\begin{align*}
&f_n=\sum_{k=0}^n(-1)^k\binom{n}{k}g_k \quad&\quad g_n=\sum_{k=0}^n(-1)^k\binom{n}{k}f_k\\
&f_n=\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2k+1} \quad&\quad\frac{1}{2n+1}=\sum_{k=0}^n(-1)^k\binom{n}{k}f_k
\end{align*}
We conclude again with (2)
\begin{align*}
\frac{1}{2n+1}&=\sum_{k=0}^n(-1)^k\binom{n}{k}f_k\\
&=\sum_{k=0}^n\frac{(-4)^{k}}{2k+1}\binom{n}{k}\binom{2k}{k}^{-1}\\
\end{align*}
This identity looks somewhat like a one-dimensional version of (1). Maybe this information can be used to solve (1).
Best Answer
We seek to evaluate
$$\sum_{l=0}^m (-4)^l {m\choose l} {2l\choose l}^{-1} \sum_{k=0}^n \frac{(-4)^k}{2k+1} {n\choose k} {2k\choose k}^{-1} {k+l\choose l}.$$
We start with the inner term and use the Beta function identity
$$\frac{1}{2k+1} {2k\choose k}^{-1} = \int_0^1 x^k (1-x)^k \; dx.$$
We obtain
$$\int_0^1 [z^l] \sum_{k=0}^n {n\choose k} (-4)^k x^k (1-x)^k \frac{1}{(1-z)^{k+1}} \; dx \\ = [z^l] \frac{1}{1-z} \int_0^1 \left(1-\frac{4x(1-x)}{1-z}\right)^n \; dx \\ = [z^l] \frac{1}{(1-z)^{n+1}} \int_0^1 ((1-2x)^2-z)^n \; dx \\ = \sum_{q=0}^l {l-q+n\choose n} [z^q] \int_0^1 ((1-2x)^2-z)^n \; dx \\ = \sum_{q=0}^l {l-q+n\choose n} {n\choose q} (-1)^q \int_0^1 (1-2x)^{2n-2q} \; dx \\ = \sum_{q=0}^l {l-q+n\choose n} {n\choose q} (-1)^q \left[-\frac{1}{2(2n-2q+1)} (1-2x)^{2n-2q+1}\right]_0^1 \\ = \sum_{q=0}^l {l-q+n\choose n} {n\choose q} (-1)^q \frac{1}{2n-2q+1}.$$
Now we have
$$ {l-q+n\choose n} {n\choose q} (-1)^q \frac{1}{2n-2q+1} \\ = \mathrm{Res}_{z=q} \frac{(-1)^n}{2n+1-2z} \prod_{p=0}^{n-1} (l+n-p-z) \prod_{p=0}^n \frac{1}{z-p}.$$
Residues sum to zero and since $\lim_{R\to\infty} 2\pi R \times R^n / R / R^{n+1} = 0$ we may evaluate the sum using the negative of the residue at $z=(2n+1)/2.$ We get
$$\frac{1}{2} (-1)^n \prod_{p=0}^{n-1} (l+n-p-(2n+1)/2) \prod_{p=0}^n \frac{1}{(2n+1)/2-p} \\ = (-1)^n \prod_{p=0}^{n-1} (2l+2n-2p-(2n+1)) \prod_{p=0}^n \frac{1}{2n+1-2p} \\ = (-1)^n \prod_{p=0}^{n-1} (2l-2p-1) \frac{2^n n!}{(2n+1)!} \\ = (-1)^n \frac{1}{2l+1} \prod_{p=-1}^{n-1} (2l-2p-1) \frac{2^n n!}{(2n+1)!} \\ = (-1)^n \frac{2^n n!}{(2n+1)!} \frac{1}{2l+1} \prod_{p=0}^{n} (2l-2p+1) \\ = (-1)^n \frac{2^{2n+1} n!}{(2n+1)!} \frac{1}{2l+1} \prod_{p=0}^{n} (l+1/2-p) \\ = (-1)^n \frac{2^{2n+1} n! (n+1)!}{(2n+1)!} \frac{1}{2l+1} {l+1/2\choose n+1}.$$
We obtain for our sum
$$(-1)^n 2^{2n+1} {2n+1\choose n}^{-1} \sum_{l=0}^m (-4)^l {m\choose l} \frac{1}{2l+1} {2l\choose l}^{-1} {l+1/2\choose n+1}.$$
We now work with the remaining sum without the factor in front. We obtain
$$\int_0^1 [z^{n+1}] \sqrt{1+z} \sum_{l=0}^m {m\choose l} (-4)^l x^l (1-x)^l (1+z)^l \; dx \\ = [z^{n+1}] \sqrt{1+z} \int_0^1 (1-4x(1-x)(1+z))^m \; dx \\ = [z^{n+1}] \sqrt{1+z} \int_0^1 \sum_{q=0}^m {m\choose q} (1-2x)^{2m-2q} (-1)^q (4x(1-x))^q z^q \; dx \\ = \sum_{q=0}^m {m\choose q} {1/2\choose n+1-q} \int_0^1 (1-2x)^{2m-2q} (-1)^q (4x(1-x))^q \; dx \\ = \sum_{q=0}^m {m\choose q} {1/2\choose n+1-q} \int_0^1 (1-2x)^{2m} \left(1-\frac{1}{(1-2x)^2}\right)^q \; dx \\ = \sum_{q=0}^m {m\choose q} {1/2\choose n+1-q} \sum_{p=0}^q {q\choose p} (-1)^p \int_0^1 (1-2x)^{2m-2p} \; dx \\ = \sum_{q=0}^m {m\choose q} {1/2\choose n+1-q} \sum_{p=0}^q {q\choose p} (-1)^p \frac{1}{2m-2p+1}.$$
Re-writing then yields
$$\sum_{p=0}^m (-1)^p \frac{1}{2m-2p+1} \sum_{q=p}^m {m\choose q} {1/2\choose n+1-q} {q\choose p}.$$
Observe that
$${m\choose q} {q\choose p} = \frac{m!}{(m-q)! \times p! \times (q-p)!} = {m\choose p} {m-p\choose m-q}$$
so that we find
$$\sum_{p=0}^m {m\choose p} (-1)^p \frac{1}{2m-2p+1} \sum_{q=p}^m {m-p\choose m-q} {1/2\choose n+1-q} \\ = \sum_{p=0}^m {m\choose p} (-1)^p \frac{1}{2m-2p+1} \sum_{q=0}^{m-p} {m-p\choose m-p-q} {1/2\choose n+1-p-q} \\ = \sum_{p=0}^m {m\choose p} (-1)^p \frac{1}{2m-2p+1} \sum_{q=0}^{m-p} {m-p\choose q} {1/2\choose n+1-p-q}.$$
Continuing we obtain
$$\sum_{p=0}^m {m\choose p} (-1)^p \frac{1}{2m-2p+1} \sum_{q=0}^{m-p} {m-p\choose q} [z^{n+1-p}] z^q \sqrt{1+z} \\ = \sum_{p=0}^m {m\choose p} (-1)^p \frac{1}{2m-2p+1} [z^{n+1-p}] \sqrt{1+z} \sum_{q=0}^{m-p} {m-p\choose q} z^q \\ = \sum_{p=0}^m {m\choose p} (-1)^p \frac{1}{2m-2p+1} [z^{n+1-p}] (1+z)^{m-p+1/2} \\ = \sum_{p=0}^m {m\choose p} (-1)^p \frac{1}{2m-2p+1} {m-p+1/2\choose n+1-p} \\ = (-1)^m \sum_{p=0}^m {m\choose p} (-1)^p \frac{1}{2p+1} {p+1/2\choose n+1-m+p} \\ = (-1)^m \sum_{p=0}^m {m\choose p} (-1)^p \frac{1}{2} \frac{1}{m-n-1/2} {p-1/2\choose n+1-m+p} \\ = (-1)^m \frac{1}{2m-2n-1} \sum_{p=0}^m {m\choose p} (-1)^p {p-1/2\choose n+1-m+p}.$$
Concluding with a closed form we establish at last
$$(-1)^m \frac{1}{2m-2n-1} \sum_{p=0}^m {m\choose p} (-1)^p [z^{n+1-m}] z^{-p} (1+z)^{p-1/2} \\ = (-1)^m \frac{1}{2m-2n-1} [z^{n+1-m}] (1+z)^{-1/2} \sum_{p=0}^m {m\choose p} (-1)^p z^{-p} (1+z)^p \\ = (-1)^m \frac{1}{2m-2n-1} [z^{n+1-m}] (1+z)^{-1/2} \left(1-\frac{1+z}{z}\right)^m \\ = \frac{1}{2m-2n-1} [z^{n+1}] (1+z)^{-1/2}.$$
We finish by re-introducing the factor in front to obtain
$$(-1)^n 2^{2n+1} {2n+1\choose n}^{-1} \frac{1}{2m-2n-1} {-1/2\choose n+1} \\ = (-1)^n 2^{2n+1} {2n+1\choose n}^{-1} \frac{1}{2m-2n-1} \frac{1}{(n+1)!} \prod_{q=0}^{n} (-1/2 -q) \\ = (-1)^n 2^{n} {2n+1\choose n}^{-1} \frac{1}{2m-2n-1} \frac{1}{(n+1)!} \prod_{q=0}^{n} (-1 -2q) \\ = 2^{n} {2n+1\choose n}^{-1} \frac{1}{2n+1-2m} \frac{1}{(n+1)!} \prod_{q=0}^{n} (1 +2q) \\ = 2^{n} {2n+1\choose n}^{-1} \frac{1}{2n+1-2m} \frac{1}{(n+1)!} \frac{(2n+1)!}{2^n n!}.$$
Yes indeed this is
$$\bbox[5px,border:2px solid #00A000]{ \frac{1}{2n+1-2m}.}$$
Here I have chosen to document the simple steps as well as the complicated ones to aid all types of readers.