<s>[INST] 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.
 [/INST] Prove by Induction:    
Base Case: n=0
0 = 0  $\checkmark$
Induction Step:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\begin{eqnarray*}\frac{1}{\binom{n+1}{m}} \sum_{k=1}^m \binom{n+1-k}{n+1-m} \frac{1}{k} &=& \frac{1}{\binom{n+1}{m}} \sum_{k=1}^m \frac{n+1-k}{n+1-m} \binom{n-k}{n-m} \frac{1}{k} \\
&=& \frac{1}{\binom{n+1}{m}} \frac{1}{n+1-m} \sum_{k=1}^m (n+1-k) \binom{n-k}{n-m} \frac{1}{k}  \\
&=& \frac{1}{\binom{n+1}{m}} \frac{1}{n+1-m} \Big[ \sum_{k=1}^m (n+1) \binom{n-k}{n-m} \frac{1}{k} -\sum_{k=1}^m \binom{n-k}{n-m} \Big] \\
&=& \frac{1}{\binom{n+1}{m}} \frac{n+1}{n+1-m}  \sum_{k=1}^m \binom{n-k}{n-m} \frac{1}{k} - \frac{1}{\binom{n+1}{m}} \frac{1}{n+1-m} \sum_{k=1}^m \binom{n-k}{n-m} \\
&=& H_n - H_{n-m} - \frac{1}{\binom{n+1}{m}} \frac{1}{n+1-m} \frac{n \binom{n-1}{n-m}}{n+1-m} \\ 
&=& H_n - H_{n-m} - \frac{m}{(n+1)(n+1-m)} \\
&=& H_n - H_{n-m} + \frac{-m+(n+1)-(n+1)}{(n+1)(n+1-m)} \\
&=& H_n - H_{n-m} + \frac{1}{n+1}-\frac{1}{n+1-m} \\
&=& \bbx{H_{n+1} - H_{n+1-m}}
\end{eqnarray*}$
</s>