TITLE: Somewhat challenging binomial identity
QUESTION [6 upvotes]: The  binomial identity  valid  for positive  integer  values $n$:
  \begin{align*}
\color{blue}{\frac{n}{4}\sum_{k=1}^n\frac{(-16)^k}{k(n+k)}\binom{n+k}{2k}\binom{2k}{k}^{-1}=\sum_{k=1}^n\frac{1}{1-2k}}\tag{1}
\end{align*}
  is somewhat difficult to  prove.  I'm trying to  show (1) using generating functions, but  wasn't successful so far. So, I'm kindly asking for support.

Some information around the problem:

We can write $\frac{n}{k(n+k)}=\frac{1}{k}-\frac{1}{n+k}$ which might help to split the left-hand side into simpler sums.
If we set $a_n(z)=\sum_{k=0}^n \binom{n+k}{2k}\binom{2k}{k}^{-1}z^k$ we have
\begin{align*}
a_n(z)+a_{n-1}(z)&=\sum_{k=0}^n\frac{2n}{n+k}\binom{n+k}{2k}\binom{2k}{k}^{-1}z^k\\
&=2\sum_{k=0}^n\left(\binom{n+k}{2k}-\frac{1}{2}\binom{n+k-1}{2k-1}\right)\binom{2k}{k}^{-1}z^k
\end{align*}
which could be useful when  we try to write (1) as telescoping  sum.
We have a representation of reciprocal binomial coefficients via the Beta function:

\begin{align*}
\binom{2k}{k}^{-1}&=(2k+1)\int_0^1z^k(1-z)^{k}\,dz\\
&=(2k+1)\int_0^1z^k\sum_{r=0}^k\binom{k}{r}(-z)^r\,dz\\
&=(2k+1)\sum_{r=0}^k\binom{k}{r}(-1)^r\int_0^1z^{k+r}\,dz\\
&=(2k+1)\sum_{r=0}^k\binom{k}{r}\frac{(-1)^r}{k+r+1}\\
&=(2k+1)\sum_{r=0}^k\binom{k}{r}\frac{(-1)^{k-r}}{2k+1-r}
\end{align*}
  which could be useful to  transform (1) into a double sum, exchange  the  sums  and try to apply some telescoping.

We also have the generating function of associated Legendre polynomials (see e.g. Binomial identities by J. Riordan):
\begin{align*}
\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{n+k}{2k}z^k\right)y^n=\frac{1-y}{1-(2+z)y+y^2}
\end{align*}

Regrettably, despite this info I wasn't able to show (1).

REPLY [2 votes]: Note: There is an interesting integral representation of (1),  namely
\begin{align*}
\frac{n}{4}\sum_{k=1}^n\frac{(-16)^k}{k(n+k)}\binom{n+k}{2k}\binom{2k}{k}^{-1}
\color{blue}{=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{\cos(2nz)-1}{\sin z}\,dz
=}\sum_{k=1}^n\frac{1}{1-2k}
\end{align*}
The validity of this equality-chain is shown in this MSE post.