<s>[INST] Writing $ \frac{1}{n}\sum_{i=0}^{n-1}{\frac{m}{m-i}} =\frac{m}{n}\left( H_n-H_{m-n} \right) $ Given the following series, where $m,n$ are integers such that $n<m$:
\begin{align}
\frac{1}{n}\sum_{i=0}^{n-1}{\frac{m}{m-i}}\\
=\frac{m}{n}\left( H_n-H_{m-n} \right) 
\end{align}
We know that harmonic number $H_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + + \frac{1}{n}$. So,
\begin{align}
\frac{1}{n}\sum_{i=0}^{n-1}{\frac{m}{m-i}}\\
=\frac{m}{n}\sum_{i=0}^{n-1}{\frac{1}{m-i}}\\
=\frac{m}{n}(\frac{1}{m-0}+\frac{1}{m-1}+\frac{1}{m-2}+\cdots \frac{1}{(m-n)+1})\\
\end{align}
So,
\begin{align}
H_n=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots \frac{1}{n}\\
H_{m-n}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots \frac{1}{m-n}\\
\end{align}
Since $m>n$, how we can guarantee as well that $H_n - H_{m-n}$ is positive please? So, I see that difference here $H_n - H_{m-n} > 0$.
Problem 1: How we can proceed please to derive $\frac{m}{n}\left( H_n-H_{m-n} \right)$?
Edit: I made a mistake as it's $H_m - H_{m-n}$.
 [/INST] 
We obtain for positive integers $m>n$
\begin{align*}
\color{blue}{\frac{1}{n}\sum_{i=0}^{n-1}\frac{m}{m-i}}
&=\frac{m}{n}\sum_{i=0}^{n-1}\frac{1}{m-i}\\
&=\frac{m}{n}\sum_{i=0}^{n-1}\frac{1}{m-(n-1-i)}\tag{1}\\
&=\frac{m}{n}\sum_{i=0}^{n-1}\frac{1}{m-n+1+i}\\
&=\frac{m}{n}\sum_{i=m-n+1}^m\frac{1}{i}\tag{2}\\
&=\frac{m}{n}\left(\sum_{i=1}^m\frac{1}{i}-\sum_{i=1}^{m-n}\frac{1}{i}\right)\tag{3}\\
&\,\,\color{blue}{=\frac{m}{n}\left(H_m-H_{m-n}\right)}
\end{align*}
and the claim follows.

Comment:

*

*In (1) we change the order of summation: $i\to n-1-i$.


*In (2) we shift the index and start with $i=m-n+1$. To compensate this shift we substitute $i$ with $i-m+n-1$ within the scope of the sum.


*In (3) we write the sum as difference of Harmonic numbers.
</s>