How to calculate convergence of this function? Given $n$, is there any easy way to calculate convergence of this summation.
$$\sum_{k=0}^\infty\dfrac{1}{^{n+k}C_n}$$
EDIT: Also I need to find at which value this series converges.
 A: We have
$${n+k\choose n}=\frac{(n+k)!}{n!k!}=\frac1{n!}(k+1)(k+2)\cdots(k+n)\sim_\infty\frac{k^n}{n!}$$
hence
$$\frac{1}{{n+k\choose n}}\sim_\infty\frac{n!}{k^n}$$
and the Riemann series 
$$\sum_{k\ge1}\frac1{k^n}$$
is convergent if and only if $n>1$.
Edit The sum of this series by telescoping is:
$$n!\sum_{k=0}^\infty\frac1{(k+1)(k+2)\cdots(k+n)}\\=\frac{n!}{n-1}\sum_{k=0}^\infty\frac1{(k+1)(k+2)\cdots(k+n-1)}-\frac1{(k+2)(k+3)\cdots(k+n)}\\=\frac{n!}{(n-1)\times (n-1)!}=\frac n{n-1}$$
A: Using the recurrence for Pascal's Triangle, we get
$$
\begin{align}
\frac1{\binom{k-1}{n-1}}-\frac1{\binom{\vphantom{1}k}{n-1}}
&=\frac{\binom{k-1}{n-2}}{\binom{k-1}{n-1}\binom{\vphantom{1}k}{n-1}}\\[4pt]
&=\frac{\frac{n-1}{n}}{\binom{\vphantom{1}k}{n}}\tag{1}
\end{align}
$$
Summing $(1)$ we get
$$
\begin{align}
\sum_{k=0}^\infty\frac1{\binom{k+n\vphantom{1}}{n}}
&=\sum_{k=n}^\infty\frac1{\binom{k\vphantom{1}}{n}}\\
&=\lim_{m\to\infty}\sum_{k=n}^m\frac1{\binom{k\vphantom{1}}{n}}\\
&=\lim_{m\to\infty}\frac{n}{n-1}\sum_{k=n}^m\left[\frac1{\binom{k-1}{n-1}}-\frac1{\binom{\vphantom{1}k}{n-1}}\right]\\
&=\lim_{m\to\infty}\frac{n}{n-1}\left[1-\frac1{\binom{\vphantom{1}m}{n-1}}\right]\tag{2}
\end{align}
$$
As long as $n\gt1$, the series converges to $\dfrac{n}{n-1}$.
A: Consider the $k(\frac {a_k}{a_{k+1}}−1)=k(\frac {n+k+1}{k+1}−1)=\frac{kn}{k+1}→n$ If $k→∞$ so the series converges iff $n>1$[By, Raabe's Test. Also see that if $n=1$ then $a_k=\frac1k$ the series is divergent].
