Show that $\sum\nolimits_{d|n} \frac{1}{d} = \frac{\sigma (n)}{n}$ for every positive integer $n$.
where $\sigma (n)$ is the sum of all the divisors of $n$
and $\sum\nolimits_{d|n} f(d)$ is the summation of $f$ at each $d$ where $d$ is the divisor of $n$.
I have written $n=p_1^{\alpha_1}p_2^{\alpha_2}p_3^{\alpha_3}…….p_k^{\alpha_k}$ then:-
$$\begin {align*} \sum\nolimits_{d|n} \frac{1}{d}&=\frac{d_2.d_3……d_m+d_1.d_3……d_m+……..+d_1.d_2.d_3……d_{m-1}}{d_1.d_2.d_3……d_m} \\&=\frac{d_2.d_3……d_m+d_1.d_3……d_m+……..+d_1.d_2.d_3……d_{m-1}}{p_1^{1+2+…+\alpha_1}p_2^{1+2+…+\alpha_2}p_3^{1+2+….+\alpha_3}…….p_k^{1+2+….+\alpha_k}} \\ \end{align*}$$
where $d_i$ is some divisor among the $m$ divisors.
Then I cannot comprehend the numerator so that to get the desired result.
Also suggest some other approches to this question.
Best Answer
$\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} $