I want to show that $$ \sum_{k=0}^n \frac{(-1)^k}{k+1} {n \choose k} = \frac{1}{n+1} $$ I know, that I will have to use that for $ l \geq 1 $ $$ \sum_{i=0}^l (-1)^i {l \choose i} = 0 $$. My idea is that I write: $$ \sum_{k=0}^n \frac{(-1)^k}{k+1} {n \choose k} = \sum_{k=0}^n (-1)^k {n \choose k} \frac{1}{k+1}$$ but in order to continue with my proof I know that I have to pull out a constant from the sum because I'm pretty sure the last step of the proof ought to be using the second "sum-rule" to make it 0 and there will be only $ \frac{1}{n+1}$ left.
Summary: my problem is that I don't know how to get to that second line (beginning at "..that for $l\geq1$") because there is a l and i and in my last line there is a n and a k.
Furthermore I don't know how to get that "k+1" to become "n+1" and whether I can just pull it out of the sum since it isn't a constant but dependent upon the k, which is running towards n.
