An inverse binomial identity If $n \in \mathbb{N}$, how can one show that:
$$\sum_{k = 1}^n \frac{(n - k)! \, n! \, k!}{(2n - 2k + 1)! \, (2k)!} = \sum_{k = 1}^n \frac{2^{k - n}}{k\binom{2k}{k}} \tag{*}$$
Ideally I would like to start with the left hand side of ($*$) and somehow arrive at the right hand side. In terms of binomial coefficients, the left hand side (LHS) of ($*$) can be written as
$$\text{LHS} = \sum_{k = 1}^n \frac{\binom{n}{k}}{\binom{2n - 2k + 1}{n - k + 1}} \frac{1}{(n - k + 1) \binom{2k}{k}},$$
but that hasn't helped me very much. Other than that, I unfortunately do not have much more to offer.
 A: Using binomial identities, the proposed equation is equivalent to proving,
$$ (1) \quad \quad \sum_{k=1}^n \frac{\binom{2n}{2k}}{\binom{n}{k}} \frac{1}{2n-2k+1}
= 2^{-n}\binom{2n}{n}\sum_{k=1}^n \frac{2^k}{k \binom{2k}{k}}$$
Use the integral identity for the reciprocal binomial,
$$ \frac{1}{\binom{n}{k}} = (n+1)\int_0^1t^k(1-t)^{n-k}. $$
Insert it into the left-hand side (LHS) of (1), interchange $\sum$ and $\int$, and do the sum explicitly.  The integral is long and hideous, but we'll simplify it to something simple.  In that integral, let $t=\sin^2(u).$ To shorten the equation, $c=\cos{u}$ and $s=\sin{u}.$ Then
$$ J_n:=\frac{2n+1}{n+1}\sum_{k=1}^n \frac{\binom{2n}{2k}}{\binom{n}{k}} \frac{1}{2n-2k+1}
= $$
$$=\int_0^{\pi/2} du \ c \Big(c \big(  (c+s)^{2n} - (c-s)^{2n} \big) + s \big(-2 \underline{s^{2n} }
+ (c-s)^{2n} + (c+s)^{2n} \big) \Big) $$
The underlined term can be solved in closed form.  The $(c \pm s)$ terms can be combined to one trig expression with an offset of $\pi/4.$  Let $P=\sin(u+\pi/4)$ and $M=\sin(u-\pi/4)$.  Then
$$J_n=\frac{-1}{n+1} + 2^n\int_0^{\pi/2} du \ c^2\big(P^{2n} - M^{2n} \big) + c\ s \big(P^{2n} + M^{2n} \big) .$$
Shift integral limit by $\pi/2$ and use double-angle trig ID's,
$$J_n=\frac{-1}{n+1} + 2^n\int_{-\pi/4}^{\pi/4} du \ \frac{1+\sin{2u}}{2}\big(\cos^{2n}u
-\sin^{2n}u \big) + \frac{\cos{2u}}{2}\big(\cos^{2n}u
+\sin^{2n}u \big) $$
Symmetry of integrand and more ID's finally lead to a nice expression,
$$J_n=\frac{-1}{n+1} + 2^{n+1}\int_0^{\pi/4} \big(\cos^{2n+2}u
-\sin^{2n+2}u \big) du .$$
Mathematica knows
$$\int_0^{\pi/4} \cos^{s}u \ du = \frac{\sqrt{\pi} \ \Gamma(1/2+s/2)}{2\Gamma(1+s/2)} - 2^{-s/2}{}_2F_1(1,1/2,3/2+s/2,-1)/(s+1)$$
and
$$\int_0^{\pi/4} \sin^{s}u \ du = 2^{-s/2}{}_2F_1(1,1/2,3/2+s/2,-1)/(s+1)$$
where the ${}_2F_1(\cdot)$ is Gauss's hypergeometric function.
Thus
$$ (2) \quad J_n=\frac{-1}{n+1} + \binom{2n+2}{n+2}2^{-(n+1)}\frac{\pi}{2} - \frac{2}{2n+3} \ 
{}_2F_1(1,1/2,n+5/2,-1)$$
Now consider the RHS of (1) with
$$a(n)=2^{-n}\binom{2n}{n}\sum_{k=1}^n \frac{2^k}{k \binom{2k}{k}}.$$
It is easy to show the recursion
$$ (3) \quad \quad a(n+1)=\frac{2n+1}{n+1}a(n) + \frac{1}{n+1} $$
With the initial condition a(1) =1, Mathematica solves it (RSolve) as
$$(4) \quad \quad a(n)=\frac{\pi}{2}2^{-n}\binom{2n}{n} - \frac{2}{2n+1} \  {}_2F_1(1,1/2,n+3/2,-1) $$
Letting $\tilde{J_n} = (2n+1)/(n+1)a(n),$ from eq. (3) we find
$$ \tilde{J_n} \overset{(3)}{=} \  a(n+1) - \frac{1}{n+1} \overset{(2,4)}{=} \ J_n .$$
This completes the proof.
A consequence is that
$$(5) \quad \int_0^{\pi/4} \big(\cos^{2n}(u) - \sin^{2n}(u) \big) \  du =
\frac{(1/2)_n}{n!}\sum_{k=1}^n \frac{k!}{(1/2)_k} \frac{2^{-k}}{k} $$
where the Pochhammer symbol has been used.  A Fourier series of the even powers of $\cos$ and $\sin$ can be inserted on the LHS of (5), integrated, and the result expressed as
$$ (6) \quad \sum_{k=1}^n \frac{k!}{(1/2)_k} \frac{2^{-k}}{k} = 2 \sum_{k=1}^n \frac{n!^2}{(n+k)!(n-k)!} \frac{\sin{(\pi k/2)}}{k} ,$$
yielding a 3rd sum to accompany the original two.
A: Hint:
rewrite the binomials / factorials through the Gamma function and apply the
duplication formula for Gamma
