If you know the calculus, there's a simple proof of your formula. However, unfortunately, I couldn't find any purely algebraic proof.
For n any non-negative integer and x any real,
$$(1+x)^n=\sum_{i=0}^nC_i^nx^i$$
Now integrate each side of the above equation from 0 to 1; first the integral of the left side:
$$\int_0^1(1+x)^n\,dx={(1+1)^{n+1}\over n+1}-{(1+0)^{n+1}\over n+1}={2^{n+1}-1\over n+1}$$
Now the integral of the right side:
$$\sum_{i=0}^nC^n_i\int_0^1x^i\,dx=\sum_{i=0}^nC^n _i{1^{i+1}-0^{i+1}\over i+1}=\sum_{i=0}^n{C^n_i\over i+1}$$