Why is $ \frac{m}{n} \sum\limits_{i=0}^{n-1} \frac{1}{m-i} = \frac{1}{n/m} \sum\limits_{k=m-n+1}^{m} \frac{1}{k} $

Question

I don't understand why $$ \frac{m}{n} \sum_{i=0}^{n-1} \frac{1}{m-i} $$ is equal to $$ \frac{1}{n/m} \sum_{k=m-n+1}^{m} \frac{1}{k} $$

Can someone explain this please?

Answer 1

To show $\dfrac mn=\dfrac 1{n/m}$, multiply by $1=\dfrac mm$.

Let $k=m-i$. When $i=0, k=m$. When $i=n-1, k=m-(n-1)=m-n+1$.

So each term in the sum $\dfrac1{m-i}$ corresponds to $\dfrac1k$;

as $i$ goes from $0$ to $n-1$, $k$ goes from $m$ to $m-n+1$.

Answer 2

Consider the term on RHS, which is the finite summation(on the left) that has been written in the reverse direction($a_1+a_2+...+a_n=a_n+a_{n-1}+...+a_1$)& the change of variables from $m-i$ to $k$.When $k=m$, it's $1/m$ which is the same as the term $\frac{1}{m-i}$ when $i=0$.Similarly, you could check the first term of the series on the left is equal to the last term of the series on the right.