How can I prove that $\frac{\sigma(n)}{n} = \sum_{(d|n)} \frac{1}{d}$ for every $n \in \mathbb{Z^{+}}$?

Question

I want to show that $\displaystyle \frac{\sigma(n)}{n} = \sum_{(d|n)} \frac{1}{d}$ for every $n \in \mathbb{Z^{+}}$.

This is essentially a basic number theory question. I am able to get to the point where I get $$ \frac{\sigma(n)}{n}= \frac{\sum_{(d|n)} d}{n} $$ but I don't know if I can do anything further.

Thanks guys!

Answer 1

$$n \sum_{d|n} \frac{1}{d}=\sum_{d|n} \frac{n}{d}=\sum_{d|n} {d} = \sigma(n) $$

Have you understood it? I have used the equality: $\sum\limits_{d|n} {f(d)}=\sum\limits_{d|n} {f(\frac{n}{d})}$.