TITLE: Proving $\sum _{k=1}^n \frac{(-1)^{k-1} 16^k (k-1)! k! (k+n-1)!}{((2 k)!)^2 (n-k)!}=\frac{4}{n}\sum _{k=1}^n \frac{1}{2 k-1}$
QUESTION [7 upvotes]: How can one prove
$$
\sum_{k = 1}^{n}\frac{\left(-1\right)^{k - 1}\, 16^{k}\,
\left(k - 1\right)!\, k!\, \left(k + n - 1\right)!}
{\left[\left(2k\right)!\right]^{\, 2}\,\left(n - k\right)!} =
\frac{4}{n}\sum_{k = 1}^{n}\frac{1}{2k - 1}
$$
I was given this without proof, but only a hint (to evaluate $\int_{0}^{\pi/2}\frac{2}{n}\,\frac{1 - \cos\left(2nx\right)}{\sin\left(x\right)} \, dx$ in $2$ ways) instead. By induction the integral is easily seen equivalent to RHS, but I wonder how on earth it's related to LHS. Any help will be appreciated.

REPLY [7 votes]: We show for $n\geq 1$ the validity of the equality chain :
\begin{align*}
\sum _{k=1}^n \frac{(-1)^{k-1} 16^k (k-1)! k! (k+n-1)!}{((2 k)!)^2 (n-k)!}=\frac{2}{n}\int_0^{\frac{\pi }{2}} \frac{1-\cos (2 n z)}{\sin (z)} \, dz=\frac{4}{n}\sum _{k=1}^n \frac{1}{2 k-1}\tag{1}
\end{align*}

We start with the left-hand side of (1). We obtain for $n\geq 1$:
  \begin{align*}
\color{blue}{\frac{2}{n}}&\color{blue}{\int_{0}^{\frac{\pi}{2}}\frac{1-\cos(2nz)}{\sin(z)}\,dz}\\
&=\frac{2}{n}\int_{0}^{\frac{\pi}{2}}\left(1-\sum_{j=0}^n(-1)^k\binom{2n}{2j}\cos^{2n-2j}(z)\sin^{2j}(z)\right)\frac{dz}{\sin(z)}\tag{2}\\
&=\frac{2}{n}\int_{0}^{\frac{\pi}{2}}\left(1-\sum_{j=0}^n(-1)^k\binom{2n}{2j}\left(1-\sin^2(z)\right)^{n-j}\sin^{2j}(z)\right)\frac{dz}{\sin(z)}\\
&=\frac{2}{n}\int_{0}^{\frac{\pi}{2}}\left(1-\sum_{j=0}^n(-1)^k\binom{2n}{2j}\sum_{k=0}^{n-j}\binom{n-j}{k}(-1)^k\sin^{2j+2k}(z)\right)\frac{dz}{\sin(z)}\\
&=\frac{2}{n}\sum_{j=0}^n\sum_{{k=0}\atop{(j,k)\ne(0,0)}}^{n-j}\binom{2n}{2j}\binom{n-j}{k}(-1)^{j+k-1}\int_{0}^{\frac{\pi}{2}}\sin^{2j+2k-1}(z)\,dz\\
&=\frac{2}{n}\sum_{j=0}^n\sum_{{k=0}\atop{(j,k)\ne(0,0)}}^{n-j}\binom{2n}{2j}\binom{n-j}{k}(-1)^{j+k-1}\frac{4^{j+k-1}}{2j+2k-1}\binom{2j+2k-2}{j+k-1}^{-1}\tag{3}\\
&=\frac{2}{n}\sum_{j=0}^n\sum_{{k=j}\atop{(j,k)\ne(0,0)}}^{n}\binom{2n}{2j}\binom{n-j}{k-j}(-1)^{k-1}\frac{4^{k-1}}{2k-1}\binom{2k-2}{k-1}^{-1}\tag{4}\\
&=\frac{2}{n}\sum_{k=1}^n\binom{2k-2}{k-1}^{-1}\frac{(-4)^{k-1}}{2k-1}\sum_{j=0}^k\binom{2n}{2j}\binom{n-j}{k-j}\tag{5}\\
&=\frac{2}{n}\sum_{k=1}^n\binom{2k-2}{k-1}^{-1}\frac{(-4)^{k-1}}{2k-1}\binom{n+k}{n-k}\frac{4^kn}{n+k}\tag{6}\\
&=\frac{1}{2}\sum_{k=1}^n\frac{(k-1)!(k-1)!}{(2k-2)!}\,\frac{(-1)^{k-1}16^k}{2k-1}\,\frac{(n+k)!}{(2k)!(n-k)!}\,\frac{1}{n+k}\\
&\,\,\color{blue}{=\sum_{k=1}^n\frac{(k-1)!k!}{(2k)!}\,\frac{(-1)^{k-1}16^k(n+k-1)!}{(2k)!(n-k)!}}
\end{align*}
  and the claim follows.

Comment:

In (2) we use the trigonometric summation identity
\begin{align*}
\cos(2nz)=\sum_{j=0}^n(-1)^k\binom{2n}{2j}\cos^{2n-2j}(z)\sin^{2j}(z)
\end{align*}
In (3) we use the identity $\int_{0}^{\frac{\pi}{2}}\sin^{2n+1}(z) dz=\frac{4^n}{2n+1}\binom{2n}{n}^{-1}$. See for instance this MSE post.
In (4) we shift the index $k$ to start with $k=j$.
In (5) we exchange the sums.
In (6) we use the binomial identity $\sum_{j=0}^k\binom{2n}{2j}\binom{n-j}{k-j}=\binom{n+k}{n-k}\frac{4^kn}{n+k}$ valid for $1\leq k\leq n$. See for instance this MSE  post.


The right-hand side of (1): We obtain
  \begin{align*}
\color{blue}{\frac{2}{n}}&\color{blue}{\int_{0}^{\frac{\pi}{2}}\frac{1-\cos(2nz)}{\sin(z)}\,dz}\\
&=\frac{2}{n}\int_{0}^{\frac{\pi}{2}}\Re\left(2i\cdot\frac{1-e^{2inz}}{e^{iz}-e^{-iz}}\right)\,dz\tag{7}\\
&=-\frac{4}{n}\int_{0}^{\frac{\pi}{2}}\Re\left(ie^{iz}\cdot\frac{e^{2inz}-1}{e^{2iz}-1}\right)\,dz\\
&=-\frac{4}{n}\int_{0}^{\frac{\pi}{2}}\Re\left(ie^{iz}\sum_{k=0}^{n-1}e^{2ikz}\right)\,dz\tag{8}\\
&=-\frac{4}{n}\Re\left(i\sum_{k=0}^{n-1}\int_{0}^{\frac{\pi}{2}}e^{(2k+1)iz}\right)\,dz\\
&=-\frac{4}{n}\Re\left(\left.\sum_{k=0}^{n-1}\frac{1}{2k+1}e^{(2k+1)iz}\right|_{0}^{\frac{\pi}{2}}\right)\\
&=\frac{4}{n}\sum_{k=0}^{n-1}\frac{1}{2k+1}\\
&\,\,\color{blue}{=\frac{4}{n}\sum_{k=1}^{n}\frac{1}{2k-1}}\tag{9}
\end{align*}
  and the claim follows.

Comment:

In (7) we use the identities $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$, $\cos(z)=\frac{e^{iz}+e^{-iz}}{2}$.
In (8) we use the finite geometric summation formula.
In (9) we shift the index to start with $k=1$.


Note: When using binomial coefficients and the harmonic numbers $H_n=\sum_{k=1}^n\frac{1}{k}$ for $n\geq 1$ and $H_0=0$ we can write the equality chain (1) for $n\geq 1$ as
  \begin{align*}
-\sum_{k=1}^n\frac{(-16)^k}{k(n+k)}\binom{n+k}{n-k}\binom{2k}{k}^{-1}
=\frac{2}{n}\int_0^{\frac{\pi }{2}} \frac{1-\cos (2 n z)}{\sin (z)} \, dz
=\frac{4}{n}\left(H_{2n-1}-\frac{1}{2}H_{n-1}\right)
\end{align*}