Show that $\sum\nolimits_{d|n} \frac{1}{d} = \frac{\sigma (n)}{n}$ for every positive integer $n$.

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

or

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

\begin{equation*} \begin{split} \sigma (n) &=\sum_{d|n}d (\text{ i.e. sum of all divisors of $n$})\\ &=\sum_{d|n}\left(\frac{n}{d}\right) (\text{ i.e. sum of all divisors of $n$})\\ \therefore \sigma(n)&= n\sum_{d|n}\frac{1}{d}\\ \Rightarrow \sum_{d|n}\frac{1}{d}=\frac{\sigma(n)}{n}. \end{split} \end{equation*}