Writing $ \frac{1}{n}\sum_{i=0}^{n-1}{\frac{m}{m-i}} =\frac{m}{n}\left( H_n-H_{m-n} \right) $

Question

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 $H_m - H_{m-n}$.

Answer 1

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.