Simplification of a double sum involving partial sums of harmonic series

Question

Could somebody explain the jump in the following equation? $$\frac{1}{n}\sum\limits_{i=1}^{n}\left[1 + \sum\limits_{j=i+1}^{n}\frac{1}{m}\right] = 1 + \frac{1}{nm}\sum\limits_{i=1}^{n}(n-i) $$

Answer 1

$$\frac{1}{n}\sum\limits_{i=1}^{n}\left[1 + \sum\limits_{j=i+1}^{n}\frac{1}{m}\right]\\ =\frac{1}{n}\sum_{i=1}^n 1 +\frac{1}{n} \sum\limits_{i=1}^{n}\sum\limits_{j=i+1}^{n}\frac{1}{m}$$

Sum of $n$ $1$'s is $n$, and $\frac{1}{m}$ is not related to the index, so we can factor it out. This gives

$$\frac{1}{n}\cdot n+\frac{1}{nm}\sum\limits_{i=1}^{n}\sum\limits_{j=i+1}^{n} 1$$

Now $\sum\limits_{j=i+1}^{n} 1$ is $(n-i)$ $1$'s. Hence the result.