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?

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{k = 1}^{n}{\pars{-1}^{k - 1} \over k}{n \choose k} & =
\sum_{k = 1}^{n}\pars{-1}^{k - 1}{n \choose k}\int_{0}^{1}t^{k - 1}\,\dd t =
-\int_{0}^{1}\sum_{k = 1}^{n}{n \choose k}\pars{-t}^{k}\,{\dd t \over t}
\\[5mm] & =
-\int_{0}^{1}{\pars{1 - t}^{n} - 1 \over t}\,\dd t =
\int_{0}^{1}{t^{n} - 1 \over t - 1}\,\dd t =
\int_{0}^{1}\sum_{k = 0}^{n - 1}t^{k}\,\dd t =
\sum_{k = 0}^{n - 1}\int_{0}^{1}t^{k}\,\dd t
\\[5mm] & =
\sum_{k = 0}^{n - 1}{1 \over k + 1}\,\dd t =
\sum_{k = 1}^{n}{1 \over k}\,\dd t =
\bbx{1 + {1 \over 2} + \cdots + {1 \over n}}
\end{align}
