How to prove that $\sum_{k=1}^{n-1}\frac{\binom{n-1}{k-1}}{k}=\frac{2^n-2}{n}$?

Question

My problem is to prove the identity:

$$\sum_{k=1}^{n-1}\frac{\binom{n-1}{k-1}}{k}=\frac{2^n-2}{n}$$

Basically I am trying to show that the LHS is equal to the RHS. I have put this into wolfram and the LHS does simplify to $\frac{2^n-2}{n}$, however, I am not sure as to how it did that.

Answer 1

Note that $$\frac 1k\binom{n-1}{k-1}=\frac 1k\cdot\frac{(n-1)!}{(n-k)!(k-1)!}=\frac{n\cdot (n-1)!}{(n-k)!k!}\cdot \frac 1n=\frac 1n\binom{n}{k}.$$ Then, use the binomial theorem : $$\sum_{k=1}^{n-1}\frac{\binom{n-1}{k-1}}{k}=\frac 1n\sum_{k=1}^{n-1}\binom{n}{k}=\frac 1n(2^n-2)$$

Answer 2

Consider that, by the binomial theorem: $$ \sum_{k=1}^{n-1}\binom{n-1}{k-1}x^{k-1} = (x+1)^{n-1}-x^{n-1}\tag{1} $$ hence by integrating both sides over $[0,1]$ we get: $$ \sum_{k=1}^{n-1}\frac{\binom{n-1}{k-1}}{k} = \frac{1}{n}\left[(x+1)^n-x^n\right]_{0}^{1}=\frac{2^n-\color{red}{2}}{n}. \tag{2}$$