Sum of infinite series $\sum_{n=50}^{\infty} \frac{1}{\binom{n}{50}}$ Find Sum of infinite series $$S=\sum_{n=50}^{\infty} \frac{1}{\binom{n}{50}}$$ My Try is :
$$50S=\sum_{n=50}^{\infty} \frac{n-(n-50)}{\binom{n}{50}}$$ so
$$50S=\sum_{n=50}^{\infty}\frac{n}{\binom{n}{50}}-\sum_{n=50}^{\infty}\frac{n-50}{\binom{n}{n-50}} $$ so
$$50S=\sum_{n=50}^{\infty}\frac{n}{\binom{n}{50}}-\sum_{n=0}^{\infty}\frac{n}{{\binom{n+50}{50}}}$$
any clue further
 A: An alternative approach through Euler's Beta function:
$$\begin{eqnarray*}\sum_{n\geq 50}\binom{n}{50}^{-1}=50\sum_{n\geq 50}\frac{\Gamma(50)\,\Gamma(n+1-50)}{\Gamma(n+1)}&=&50\sum_{n\geq 50}B(50,n+1-50)\\&=&50\int_{0}^{1}\sum_{n\geq 50}(1-u)^{49}u^{n-50}\,du\\&=&50\int_{0}^{1}(1-u)^{48}\,du = \color{red}{\frac{50}{49}}.\end{eqnarray*} $$
A: I dont know exactly now how to proceed with standard manipulations but from finite calculus we have that
$$\sum k^\underline h\delta k=\frac{k^{\underline h+1}}{h+1}+C$$
And $$k^{\underline {-h}}=\frac1{(k+1)^{\overline h}},\quad h\neq 1$$ 
Hence
$$\begin{align}\sum_{k=50}^\infty \binom{k}{50}^{-1}&=50! \sum\nolimits_{50}^\infty\frac{\delta k}{k^{\underline {50}}}=50! \sum\nolimits_{50}^\infty\frac{\delta k}{(k-50+1)^{\overline {50}}}\\&=50!\sum\nolimits_{50}^\infty (k-50)^{\underline{-50}}=50!\frac{(k-50)^{\underline{-49}}}{-49}\bigg|_{50}^\infty\\&=\frac{50}{49}\end{align}$$
A: Hint:
$$ \frac{1}{\binom{n}{50}} = \frac{50}{49} \bigg( \frac{1}{\binom{n-1}{49}} - \frac{1}{\binom{n}{49}} \bigg). $$
