Binomial Harmonic Numbers Prove this equation for $0 \leq m \leq n$:
$$
\frac{1}{\binom{n}{m}}\sum_{k=1}^m \binom{n-k}{n-m} \frac{1}{k} = H_n - H_{n-m}
$$
where $H_k$ denotes the k-th harmonic number $\left(~H_k := \sum_{n=1}^k \frac{1}{n}~\right)$.  
Tried to use Abels partial summation $\big(\sum_{k=1}^m a_k b_k = a_m \sum_{k=1}^m - \sum_{k=1}^{m-1} (a_{k+1}-a_k)\sum_{i=1}^k b_i \big)$, but it leads to nowhere.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{{1 \over {n \choose m}}\sum_{k = 1}^{m}{n - k \choose n - m}{1 \over k} = H_{n} - H_{n - m}:\ {\Large ?}.\quad H_{z}}$ is a Harmonic Number.
  $\ds{\quad0 \leq m \leq n}$.

\begin{align}
{1 \over {n \choose m}}\sum_{k = 1}^{m}{n - k \choose n - m}{1 \over k} & =
{1 \over {n \choose m}}\sum_{k = 1}^{\infty}{n - k \choose m - k}{1 \over k} =
{1 \over {n \choose m}}\sum_{k = 1}^{\infty}{m - n - 1 \choose m - k}
\pars{-1}^{m - k}\,{1 \over k}
\\[5mm] & =
{\pars{-1}^{m} \over {n \choose m}}\sum_{k = 1}^{\infty}
\braces{\bracks{z^{m - k}}\pars{1 + z}^{m - n - 1}}
\pars{-1}^{k}\int_{0}^{1}t^{k - 1}\,\dd t
\\[5mm] & =
{\pars{-1}^{m} \over {n \choose m}}\bracks{z^{m}}
\pars{1 + z}^{m - n - 1}\int_{0}^{1}\sum_{k = 1}^{\infty}\pars{-tz}^{k}
\,{\dd t \over t}
\\[5mm] & =
{\pars{-1}^{m} \over {n \choose m}}\bracks{z^{m}}
\pars{1 + z}^{m - n - 1}\int_{0}^{1}{-tz \over 1 + tz}\,{\dd t \over t}
\\[5mm] & =
{\pars{-1}^{m + 1} \over {n \choose m}}\bracks{z^{m}}
\pars{1 + z}^{m - n - 1}\ln\pars{1 + z}
\\[5mm] & =
{\pars{-1}^{m + 1} \over {n \choose m}}
\left.\partiald{}{\epsilon}\bracks{z^{m}}
\pars{1 + z}^{m - n - 1 + \epsilon}\,\right\vert_{\ \epsilon\ =\ 0}
\\[5mm] & =
{\pars{-1}^{m + 1} \over {n \choose m}}
\left.\partiald{}{\epsilon}{m - n - 1 + \epsilon \choose m}
\,\right\vert_{\ \epsilon\ =\ 0}
\\[5mm] & =
{\pars{-1}^{m + 1} \over {n \choose m}}
\left.\partiald{}{\epsilon}{n - \epsilon \choose m}\pars{-1}^{m}
\,\right\vert_{\ \epsilon\ =\ 0}
\\[5mm] & =
{\pars{-1}^{m + 1} \over {n \choose m}}\braces{-{n - \epsilon \choose m}
\bracks{H_{n - \epsilon} - H_{n - m - \epsilon}}\pars{-1}^{m}}
_{\ \epsilon\ =\ 0}
\\[5mm] &=
\bbx{H_{n} - H_{n - m}}
\end{align}
A: Prove by Induction:    
Base Case: n=0
0 = 0  $\checkmark$
Induction Step:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\begin{eqnarray*}\frac{1}{\binom{n+1}{m}} \sum_{k=1}^m \binom{n+1-k}{n+1-m} \frac{1}{k} &=& \frac{1}{\binom{n+1}{m}} \sum_{k=1}^m \frac{n+1-k}{n+1-m} \binom{n-k}{n-m} \frac{1}{k} \\
&=& \frac{1}{\binom{n+1}{m}} \frac{1}{n+1-m} \sum_{k=1}^m (n+1-k) \binom{n-k}{n-m} \frac{1}{k}  \\
&=& \frac{1}{\binom{n+1}{m}} \frac{1}{n+1-m} \Big[ \sum_{k=1}^m (n+1) \binom{n-k}{n-m} \frac{1}{k} -\sum_{k=1}^m \binom{n-k}{n-m} \Big] \\
&=& \frac{1}{\binom{n+1}{m}} \frac{n+1}{n+1-m}  \sum_{k=1}^m \binom{n-k}{n-m} \frac{1}{k} - \frac{1}{\binom{n+1}{m}} \frac{1}{n+1-m} \sum_{k=1}^m \binom{n-k}{n-m} \\
&=& H_n - H_{n-m} - \frac{1}{\binom{n+1}{m}} \frac{1}{n+1-m} \frac{n \binom{n-1}{n-m}}{n+1-m} \\ 
&=& H_n - H_{n-m} - \frac{m}{(n+1)(n+1-m)} \\
&=& H_n - H_{n-m} + \frac{-m+(n+1)-(n+1)}{(n+1)(n+1-m)} \\
&=& H_n - H_{n-m} + \frac{1}{n+1}-\frac{1}{n+1-m} \\
&=& \bbx{H_{n+1} - H_{n+1-m}}
\end{eqnarray*}$
A: This solution is similar to an inductive proof with respect to $m$.
Let the sum of interest be
$$s_{n,m}=\frac{1}{\binom{n}{m}} \sum _{k=1}^m \frac{1}{k} \binom{n-k}{n-m}$$
For $m=1$ we find
$$s_{n,1}=\frac{1}{n}$$
The difference of $s$ with respect to $m$ turns out to be surprisingly simple:
$$s_{n,{m+1}}-s_{n,m}=\frac{1}{n-m}\tag{1}$$
so that
$$s_{n,2}=\frac{1}{n}+\frac{1}{n-1}$$
$$s_{n,3}=\frac{1}{n}+\frac{1}{n-1}+\frac{1}{n-2}$$
$$...$$
$$s_{n,m}=\frac{1}{n}+\frac{1}{n-1}+\frac{1}{n-2}+...+\frac{1}{n-m+1}$$
The last expression is easily identified as the difference between two harmonic numbers. Hence
$$s_{n,m}=H_{n}-H_{n-m}$$
as requested.
And now we prove the recurence relation (1) using the definition
$$\binom{n}{m}=\frac{n!}{m!\;(n-m)!}$$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\begin{eqnarray*}
s_{n,{m+1}}-s_{n,m} 
&=& \frac{1}{\binom{n}{m+1}} \sum_{k=1}^{m+1} \frac{1}{k}\binom{n-k}{n-m-1} - \frac{1}{\binom{n}{m}} \sum_{k=1}^{m} \frac{1}{k}\binom{n-k}{n-m}\\
&=& \frac{1}{\binom{n}{m}}\left( \frac{m+1}{n-m}\left( \sum_{k=1}^m \frac{1}{k} \binom{n-k}{n-m-1} +\frac{1}{m+1}\right) -\sum_{k=1}^m \frac{1}{k} \binom{n-k}{n-m}\right) \\
&=& \frac{1}{\binom{n}{m}}\left( \frac{1}{n-m}+ \sum_{k=1}^m \frac{1}{k}\binom{n-k}{n-m} \left( \frac{m+1}{m+1-k} -1 \right) \right) \\
&=& \frac{1}{\binom{n}{m}}\left( \frac{1}{n-m}+ \sum_{k=1}^m \frac{1}{m+1-k}\binom{n-k}{n-m} \right) \\
&=& \frac{1}{n-m}\frac{1}{\binom{n}{m}}\left( 1+\sum_{k=1}^m \frac{n-m}{m+1-k}\binom{n-k}{n-m} \right) \\
&=& \frac{1}{n-m}\frac{1}{\binom{n}{m}}\left( 1+\sum_{k=1}^m \binom{n-k}{m-k+1} \right) =\frac{1}{n-m} \\
\end{eqnarray*}$
In the last step we have used the identity
$\begin{eqnarray*}
\binom{n}{m}
&=&  \sum_{k=1}^m \binom{n-k}{m-k+1} +1 = \sum_{k=1}^{m+1} \binom{n-k}{m-k+1} \\
\end{eqnarray*}$
which in turn results from iterating the basic recursion of the binomial coefficients
$\begin{eqnarray*}
\binom{n}{m}
&=& \binom{n-1}{m}+\binom{n-1}{m-1} \\
&=& \binom{n-1}{m}+\binom{n-2}{m-1}+\binom{n-2}{m-2} \\
&=& \binom{n-1}{m}+\binom{n-2}{m-1}+...+\binom{n-m}{1} +\binom{n-m-1}{0}\tag{2}\\
\end{eqnarray*}$
This completes the proof of (1) and hence the solution.
Comments
1) I feel that a much shorter proof should be possible starting from (2), but I didn't find it.   
2) Interesting related formula (Derivative of binomial coefficients)
$$\frac{d}{dn}\ln\binom{n}{k} = H_{n} - H_{n-k}$$
