How can i prove this? (infinite summation) I have to prove the following
$$\sum_{k=1}^n \frac{(-1)^{k-1}}{k} {n \choose k} =1+\frac{1}{2}+...+\frac{1}{n}$$
I try to prove it using that $$\sum_{k=0}^n {(-1)^{n-k}} {n \choose k} k^m = \begin{cases}
0,  & \text{if $m<n$ } \\
n!, & \text{if $m=n$ }
\end{cases} $$ 
but my biggest problem is $ (-1)^{n-k} $
Any thoughts on that or another way to approach this?

An overkill. By the Melzak's identity with $f\equiv1$ we have $$\sum_{k=1}^{n}\dbinom{n}{k}\frac{\left(-1\right)^{k-1}}{x+k}=\frac{1}{x}-\frac{1}{x\dbinom{x+n}{n}}=\frac{\dbinom{x+n}{n}-1}{x\dbinom{x+n}{n}}$$ then taking $x\rightarrow0$ and recalling that $$\frac{d}{dx}\dbinom{x+n}{n}=\dbinom{x+n}{n}\left(\psi^{\left(0\right)}\left(n+x+1\right)-\psi^{\left(0\right)}\left(x+1\right)\right)$$ where $\psi^{\left(0\right)}\left(x\right)$ is the Digamma function, we have $$\sum_{k=1}^{n}\dbinom{n}{k}\frac{\left(-1\right)^{k-1}}{k}=\color{red}{\sum_{m=1}^{n}\frac{1}{m}}.$$
