# sum of reciprocal numbers of combinations

Let $$^nC_k:=\dfrac{n!}{k!(n-k)!}$$ Please prove that,for all natural number $$k≥2$$, $$\displaystyle\sum_{n＝k+1}^{\infty}\frac{1}{^nC_k}＝\frac{1}{k-1}$$

I tried to prove by induction, but I cannot. I guess it is proved by using Tayler series for some function, but I cannot find the function.

This is very simple to prove. All you need to note is the following identity : $$\frac 1{a(a+1)...(a+m)} = \frac{1}{m}\left( \frac 1{a(a+1)...(a+m-1)} - \frac 1{(a+1)...(a+m)}\right)$$
Now, consider the sum: $$\sum_{n=k+1}^{k+m} \frac 1{\binom nk} = k! \times \sum_{i=1}^{i=m} \frac{i!}{(k+i)!} \\ = k! \sum_{i=1}^{i=m}\frac 1{(i+1)(i+2)...(i+k)} \\ = k! \sum_{i=1}^{i=m} \frac{1}{k-1}\left(\frac{1}{(i+1)...(i+k-1)} - \frac 1{(i+2)...(i+k)}\right) \\ = \frac{k!}{k-1}\sum_{i=1}^{i=m} \left(\frac{1}{(i+1)...(i+k-1)} - \frac 1{(i+2)...(i+k)}\right) \\ = \frac{k!}{k-1} \left(\frac 1{k!} - \frac 1{(m+2)...(m+k)}\right) \\ = \frac 1{k-1} - \frac{k!}{(k-1)(m+2)...(m+k)}$$
By letting $$m$$ go to infinity, the answer is clearly $$\frac 1{k-1}$$, since the second term goes to zero.