OK, so I was kinda doodling stuff in my free time and I came up with these two functions:
$$C_1(n) = \sum^{\infty}_{k = n} {{k \choose n}^{-1}}$$ $$C_2(n) = \sum^{n}_{k = 0} {{n \choose k}^{-1}}$$
I don't have Mathematica or anything at the moment, so I can't analyse these functions as such. Would anyone help me out with the properties of these functions (convergence etc.)?
The first series is divergent for $\,n\le 1\;$ and convergent for larger values to : $$C_1(n)=\sum^{\infty}_{k = n} \frac {n!}{\frac{k!}{(k-n)!}}=\sum^{\infty}_{j= 1}\frac{n!}{\frac{(j+n-1)!}{(j-1)!}}=\sum^{\infty}_{j= 1}\frac{n!}{(j)_n}=\frac {n!}{(n-1)(n-1)!}$$ where $\;(j)_n=\dfrac {(j+n-1)!}{(n-1)!}$ is the 'rising' Pochhammer symbol and where we used the equation $(19)$ from the link at the limit $n\to\infty\;$ to get : $\;\displaystyle\sum_{k=1}^\infty \frac 1{(k)_p}=\frac 1{(p-1)(p-1)!}$.
The result is simply :$$\boxed{\displaystyle C_1(n)=\frac n{n-1},\quad\text{for }\ n>1}$$
A generating function for $C_2$ is (from OEIS A046825) : \begin{align} \sum_{n=0}^\infty C_2(n)\;z^n&=\left[\frac {2\,\ln(1 - z)}{z-2}\right]'=\frac 2{(z-1)(z-2)}-\frac{2\;\ln(1-z)}{(z-2)^2}\\ &=1+2\;z+\frac 52z^2+\frac 83z^3+\frac 83z^4+\frac {13}5z^5+\frac{151}{60}z^6+\cdots\\ \end{align}
