Magic sums with binomial coefficients (PART II) In these two questions (Question 1, Question 2) we proved that for a fixed $m\geq 1$ we have
$$a_n:=\sum_{k=1}^n\frac {(2n-k-1)!\cdot k}{n!(n-k)!} {{m+k}\choose{m}} =\frac{(m+1)(m+2n)!}{n!(m+n+1)!}$$
and
$$\sum_{n=0}^\infty a_n(1/4)^n=\sum_{n=0}^\infty\frac {(m+1)(m+2n)!}{n!(m+n+1)!}(1/4)^n=2^{m+1}.$$
I was wondering if there is a direct (and easier) proof of the fact that
$$\sum_{n=0}^\infty\sum_{k=1}^n\frac {(2n-k-1)!\cdot k}{n!(n-k)!} {{m+k}\choose{m}}(1/4)^n=2^{m+1}.$$
I know that defining $a_{n,k}=\frac {(2n-k-1)!\cdot k}{n!(n-k)!}$ then $\sum_{n,k\geq 0}a_{n,k}z^nx^k=\frac{2}{x\sqrt{1-4z}-x+2}.$ Can this help to show that $\sum_{n=0}^\infty\sum_{k=1}^na_{n,k} {{m+k}\choose{m}}(1/4)^n=2^{m+1}$?
 A: Two remarks seem to be required. Firstly, your expression for $a_n$ is ill-defined for $n=0$. Indeed by all rules of summation the left side of the equation is $0$ whereas the right side is $1$. With this in mind the equation which should be proved is:
$$\sum_{n=1}^\infty\sum_{k=1}^n\frac {(2n-k-1)!\cdot k}{n!(n-k)!} {{m+k}\choose{m}}\left(\frac{1}{4}\right)^n=2^{m+1}-1.$$
Secondly, the idea to prove the identity without intermediate steps is probably not the best one because instead of several interesting results only one is obtained. And indeed it is possible to find another interesting identity hidden in the expression:
$$
\sum_{n=k}^\infty\frac{1}{n}\binom{2n-k-1}{n-1}\left(\frac{x}{4}\right)^n
=\frac{1}{k}\left(\frac{1-\sqrt{1-x}}{2}\right)^k,\tag{1}
$$ 
where $k>0$ and $|x|\le1$ are assumed.
Differentiating the expression over $x$ and shifting indices one obtains:
$$
\sum_{n=k}^\infty\binom{2n-k}{n}\left(\frac{x}{4}\right)^n
=\frac{1}{\sqrt{1-x}}\left(\frac{1-\sqrt{1-x}}{2}\right)^k,\tag{2}
$$
which can be recognized as another form of the identity proved elsewhere.
To see it:
$$
\sum_{n=k}^\infty\binom{2n-k}{n}\left(\frac{x}{4}\right)^n
=\sum_{n=k}^\infty\binom{2n-k}{n-k}\left(\frac{\sqrt{x}}{2}\right)^{2n}
=\sum_{n'=0}^\infty\binom{2n'+k}{n'}\left(\frac{\sqrt{x}}{2}\right)^{2n'+2k}\\
=\left(\frac{\sqrt{x}}{2}\right)^{k}
\sum_{n'=0}^\infty\binom{2n'+k}{n'}\left(\frac{\sqrt{x}}{2}\right)^{2n'+k}
\stackrel{*}{=}\left(\frac{\sqrt{x}}{2}\right)^{k}
\frac{1}{\sqrt{1-x}}\left(\frac{1-\sqrt{1-x}}{\sqrt{x}}\right)^k
=\frac{1}{\sqrt{1-x}}\left(\frac{1-\sqrt{1-x}}{2}\right)^k,
$$
where in $\stackrel{*}{=}$ the aforementioned identity was used. Thus (2) and consequently (1) are proved.
Substituting $x=1$ in (1) one obtains:
$$
\sum_{n=k}^\infty\frac{1}{n}\binom{2n-k-1}{n-1}\left(\frac{1}{4}\right)^n
=\frac{1}{k}\left(\frac{1}{2}\right)^k,\tag{3}
$$
The rest is straightforward:
$$\sum_{n=1}^\infty\sum_{k=1}^n\frac {(2n-k-1)!\cdot k}{n!(n-k)!} {{m+k}\choose{m}}\left(\frac{1}{4}\right)^n
=\sum_{k=1}^\infty k{{m+k}\choose{m}}\sum_{n=k}^\infty\frac {(2n-k-1)!}{n!(n-k)!}\left(\frac{1}{4}\right)^n\\
\stackrel{(3)}{=}\sum_{k=1}^\infty {{m+k}\choose{m}}\left(\frac{1}{2}\right)^k
=2^{m+1}-1,$$
as required.
The last equality can be proved in various ways, as for example by induction. Indeed for $m=0$ the equality is obvious. Assume it is valid for $m-1$:
$$
S_m:=\sum_{k=1}^\infty {{m+k}\choose{m}}\left(\frac{1}{2}\right)^k=
\sum_{k=1}^\infty \left[{{m+k-1}\choose{m}}+{{m+k-1}\choose{m-1}}\right]
\left(\frac{1}{2}\right)^k
=\frac{1}{2}\left(1+S_m\right)+S_{m-1}\\
\Rightarrow S_m=2S_{m-1}+1
\stackrel{I.H.}{=}2(2^m-1)+1=2^{m+1}-1.
$$
