An expansion of $\pi \cot(\pi/k)/(k-2)$. Mathematica suggests that for $k>=3$,
$$
\sum _{r=0}^{\infty } \frac{\left(\frac{2}{k}\right)_r}{(r+1) \left(1+\frac{1}{k}\right)_r}
=
\frac{\pi  \cot \left(\frac{\pi }{k}\right)}{k-2}
,
$$
where $(x)_r=x(x+1)\dots(x+r-1)$.
A quick look of DLMF does not find any expansion of $\cot(x)$ in this form. Any idea where this identity comes from?

Update, I did a bit more digging. It turns out
\begin{align}
\frac{1}{1-\frac{1}{k}}
\, _3F_2\left(1,\frac{1}{k},1-\frac{1}{k};1+\frac{1}{k},2-\frac{1}{k};1\right)
=
\frac{\pi  \cot \left(\frac{\pi }{k}\right)}{k-2}
.
\end{align}
The last equality id due to 
$$
\, _3F_2(1,b,1-b;b+1,2-b;1)=\frac{(\pi  b (1-b)) \cot (\pi  b)}{1-2 b}.
$$
See here.
In other words, we need to show
$$
\sum _{r=0}^{\infty } \frac{\left(\frac{2}{k}\right)_r}{(r+1) \left(1+\frac{1}{k}\right)_r}
=
\frac{1}{1-\frac{1}{k}}
\, _3F_2\left(1,\frac{1}{k},1-\frac{1}{k};1+\frac{1}{k},2-\frac{1}{k};1\right)
.
$$

Update. I have solved this one.
$$
\sum _{r=0}^{\infty } \frac{\left(\frac{2}{k}\right)_r}{(r+1) \left(1+\frac{1}{k}\right)_r} = 
\, _3F_2 (1,2/k,1;1+1/k,2;1).
$$
Then we can apply Watson's sum
$$
{{}_{3}F_{2}}\left({a,b,c\atop\frac{1}{2}(a+b+1),2c};1\right)=\frac{\Gamma%
\left(\frac{1}{2}\right)\Gamma\left(c+\frac{1}{2}\right)\Gamma\left(\frac{1}{2%
}(a+b+1)\right)\Gamma\left(c+\frac{1}{2}(1-a-b)\right)}{\Gamma\left(\frac{1}{2%
}(a+1)\right)\Gamma\left(\frac{1}{2}(b+1)\right)\Gamma\left(c+\frac{1}{2}(1-a)%
\right)\Gamma\left(c+\frac{1}{2}(1-b)\right)},
$$
See here.

With Euler’s integral transform we get :
$$\begin{eqnarray} \sum\limits_{n=0}^\infty \frac{(2x)_n}{(n+1)(1+x)_n} & = & {_3 F_2}(1,1,2x;2,x+1;1) \\ & = &\frac{\Gamma(x+1)}{\Gamma(2x)\Gamma(1-x)} \int\limits_0^1 t^{2x-1} (1-t)^{-x} {_2 F_1}(1,1;2;t)dt \\ & = & \frac{\Gamma(x+1)}{\Gamma(2x)\Gamma(1-x)} \int\limits_0^1 t^{2x-1} (1-t)^{-x} \frac{-\ln(1-t)}{t}dt \\ & = & \frac{\Gamma(x)}{\Gamma(2x)\Gamma(1-x)}\frac{d}{dz} \int\limits_0^1 t^{2x-2} (1-t)^{-xz}dt |_{z=1} \\ & = & \frac{\Gamma(x)}{\Gamma(2x)\Gamma(1-x)} \Gamma(2x-1)\frac{x\Gamma(1-x)(\psi(x)-\psi(1-x))}{\Gamma(x)} \\ & = & \frac{\pi x\cot(\pi x)}{1-2x} \end{eqnarray}$$
Where we used the reflection formula for the digamma function $\psi$ in the last step. With $\,\displaystyle x:=\frac{1}{k}\,$ your question is answered. 
