I have solved $\sum_{k = 0}^{n} \binom{n}{k} \frac{1}{k+1} = \frac{2^{n+1}-1}{n+1}$ using integrals as follows: $$(1+x)^n = \sum_{k=0}^{n}\binom{n}{k}x^k.$$$$\frac{(2)^{n+1}}{n+1} - \frac{1}{n+1} = \sum_{k=0}^{n} \binom{n}{k} \frac{1^{k+1}}{k+1}$$$$\frac{(2)^{n+1} - 1}{n+1} = \sum_{k=0}^{n} \binom{n}{k} \frac{1}{k+1} $$ (set x = 1)
However, the question states at the end "Can you also find a combinatorial proof for this identity?" I belive the answer is yes, but I am having trouble going about showing it.
