Prove $\sum_{k=1}^n \frac{(-1)^{k-1}}{k }\binom{n}{k}= H_n$ without the Beta function I know how to prove
$$\sum_{k=1}^n \frac{(-1)^{k-1}}{k }\binom{n}{k}= H_n$$
by tackling it with the beta function.
I was actually wondering if there is a proof of this fact without using the property of the Beta function $$B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
 A: 
We have
  \begin{align}
f_n&=\color{blue}{\sum_{k=1}^{n}(-1)^{k-1}\binom{n}{k}\frac{1}{k}}\\
&=\sum_{k=1}^{n}(-1)^{k-1}\left(\binom{n-1}{k}+\binom{n-1}{k-1}\right)\frac{1}{k}\\
&=f_{n-1}+\sum_{k=1}^{n}(-1)^{k-1}\binom{n-1}{k-1}\frac{1}{k}\\
&=f_{n-1}-\frac{1}{n}\sum_{k=1}^{n}(-1)^k\binom{n}{k}\\
&=f_{n-1}-\frac{1}{n}\left((1-1)^n-1\right)\\
&=f_{n-1}+\frac{1}{n}\\
&\,\,\color{blue}{=H_n}
\end{align}

Note: This can be found for instance as Example 3, section 1.2 in Combinatorial Identities by John Riordan.


Another approach is based upon generating functions and the Euler transform of a generating series $A(z)$ which is given as
  \begin{align*}
A(z)=\sum_{n=  0}^\infty a_nz^n&\quad\longrightarrow\quad
\frac{1}{1-z}A\left(\frac{z}{1-z}\right)=\sum_{n= 0}^\infty \left(\sum_{k=0}^n\binom{n}{k}a_k\right)z^n\tag{1}\\
a_n\quad&\quad\longrightarrow\quad\qquad
\sum_{k=0}^n\binom{n}{k}a_k&
\end{align*}

Applying the Euler transform (1) to the left-hand side of the identity
\begin{align*}
\sum_{k=1}^n \frac{(-1)^{k-1}}{k }\binom{n}{k}= H_n
\end{align*}
we set $a_k=\frac{(-1)^{k-1}}{k}$ and obtain as generating function
\begin{align*}
A(z)=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}z^k=\ln(1+z)\tag{2}
\end{align*}

The Euler transform of (2) is
  \begin{align*}
\frac{1}{1-z}\ln\left(1+\frac{z}{1-z}\right)&=\sum_{n=1}^\infty\left(\color{blue}{\sum_{k=1}^n\binom{n}{k}\frac{(-1)^{k-1}}{k}}\right)z^n
\end{align*}
  On the other hand we have
  \begin{align*}
\frac{1}{1-z}\ln\left(1+\frac{z}{1-z}\right)&=\frac{1}{1-z}\ln\left(\frac{1}{1-z}\right)\\
&=-\frac{1}{1-z}\ln(1-z)\\
&=\left(\sum_{k=0}^\infty z^k\right)\left( \sum_{l=1}^\infty \frac{1}{l}z^l\right)\\
&=\sum_{n=1}^\infty\left(\sum_{{k+l=1}\atop{k\geq 0,l\geq 1}}^n \frac{1}{l}\right)z^n\\
&=\sum_{n=1}^\infty\sum_{k=1}^n\frac{1}{k}z^n\\
&=\sum_{n=1}^\infty\color{blue}{H_n}z^n\\
\end{align*}
  and the claim follows.

Note: A proof of the Euler transformation formula can be found e.g.  in Harmonic Number Identities Via Euler's transform by K.N. Boyadzhiev.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{k = 1}^{n}{\pars{-1}^{k - 1} \over k}{n \choose k} = H_{n}:\ {\LARGE ?}.\quad H_{z}:\ Harmonic\ Number}$.

\begin{align}
&\bbox[10px,#ffd]{\sum_{k = 1}^{n}{\pars{-1}^{k - 1} \over k}{n \choose k}} =
\sum_{k = 1}^{\infty}{\pars{-1}^{k - 1} \over k}{n \choose n - k} =
\sum_{k = 1}^{\infty}{\pars{-1}^{k - 1} \over k}\bracks{z^{n - k}}
\pars{1 + z}^{n}
\\[5mm] = &\
\bracks{z^{n - 1}}
\pars{1 + z}^{n}\sum_{k = 1}^{\infty}{\pars{-z}^{k - 1} \over k}
=
\bracks{z^{n - 1}}\pars{1 + z}^{n}\sum_{k = 1}^{\infty}
\pars{-z}^{k - 1}\int_{0}^{1}t^{k - 1}\,\dd t
\\[5mm] = &\
\bracks{z^{n - 1}}\pars{1 + z}^{n}\int_{0}^{1}
\sum_{k = 1}^{\infty} \pars{-zt}^{k - 1}\,\dd t =
\bracks{z^{n - 1}}\pars{1 + z}^{n}\int_{0}^{1}
{1 \over 1 - \pars{-zt}}\,\dd t
\\[5mm] = &\
\bracks{z^{n - 1}}\pars{1 + z}^{n}\,{\ln\pars{1 + z} \over z} =
\bracks{z^{n}}\,
\lim_{\nu\ \to\ n}\partiald{\pars{1 + z}^{\nu}}{\nu} =
\lim_{\nu\ \to\ n}\partiald{}{\nu}{\nu \choose n}
\\[5mm] = &\
{1 \over n!}\lim_{\nu\ \to\ n}\partiald{}{\nu}
\bracks{\Gamma\pars{\nu + 1} \over \Gamma\pars{\nu - n + 1}}
\\[5mm] = &\
{1 \over n!}{\Gamma\pars{n + 1}\Psi\pars{n + 1}\Gamma\pars{1} -
\Gamma\pars{1}\Psi\pars{1}\Gamma\pars{n + 1} \over \Gamma^{2}\pars{1}}
\\[5mm] = &\ \Psi\pars{n + 1} + \gamma = \bbx{H_{n}}
\end{align}
