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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.

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$\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} $

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  • $\begingroup$ nice answer I was attempting the question from just the opposite direction. $\endgroup$ – Saurabh Jun 12 '12 at 17:39
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\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*}

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  • $\begingroup$ Is the sum of all divisors of n divided by n < (1 +1/2+ 1/3+..+ 1/n)/2 ? $\endgroup$ – 201044 Jun 24 '15 at 16:38

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