Why is $\sum_{i=1}^n\frac n{n-i+1}=n\sum_{i=1}^n\frac1i$?

Question

While reading through my (algorithms and) probability script, I have seen this equality for calculating the first moment of the coupon-collector problem. However, I don't quite see how the sum of the fraction on the left can be split up in such a way, that the sum of the fraction on the right is equal to it. $$\sum_{i=1}^n\frac n{n-i+1}=n\sum_{i=1}^n\frac1i$$ I assume it is about partial fraction decomposition, but I don't know how to apply it here.

Answer 1

It's just reverse summation order:

$$\sum_{i=1}^n \frac1i = 1 + \frac12 + \frac13 + ... + \frac1n$$ $$\sum_{i=1}^n \frac1{n-i+1} = \frac1n + \frac1{n-1} + \frac1{n-2} + ... + 1.$$

Answer 2

Pull $n$ out of the sum and make the chage of variable $j=n-i+1$.

Answer 3

This is just reindexing. As $i$ runs from $1$ to $n$, $n-i+1$ runs from $n$ to $1$. So we may rewrite the LHS as $\sum_{i=1}^n\frac ni$.