Prove that $\sum_{d\mid n}\dfrac{1}{d}=\dfrac{\sigma(n)}{n}\ \forall\ n\geq 1, n\in \mathbb Z$
My question and my approach is a lot similar to this question and a bit different from this question but just a bit short, I want to verify my approach. Here it goes:
My Approach:
Suppose $\sum_{d\mid n}\dfrac{1}{d}=\dfrac{1}{d_1}+\dfrac{1}{d_2}+\cdots+\dfrac{1}{d_r}$
Thus $\sum_{d\mid n}\dfrac{1}{d}=\dfrac{\prod_{d\mid n,\space d\neq d_1}d+\prod_{d\mid n,\space d\neq d_2}d+\cdots+\prod_{d\mid n,\space d\neq d_r}d}{\prod_{d\mid n}d}$
Now $\prod_{d\mid n}d=n^{\tau(n)/2}$ and $\prod_{d\mid n,\space d\neq d_i}d=d_jn^{\tau(n)/2-1}$ where $d_id_j=n$
This means that the numerator equals to $(d_1+d_2+\ldots+d_r)n^{\tau(n)/2-1}$ or $\sigma(n)n^{\tau(n)/2-1}$
$\therefore \sum_{d\mid n}\dfrac{1}{d}=\dfrac{\sigma(n)n^{\tau(n)/2-1}}{n^{\tau(n)/2}}=\dfrac{\sigma(n)}{n}$
For the special case when $n$ is a perfect square, say $n=m^2$, then $n^{\tau(n)/2}$ can be written as $m^{\tau(n)}$ and $n^{\tau(n)/2-1}$ can be written as $m^{\tau(n)-2}$.
Please check my method. I am also trying to search for a more simple method than this. Thus any suggestions would be welcome.
THANKS