Prove that $\sigma(n)/n = \sum_{d\mid n} 1/d$

Question

Prove that $$\frac{\sigma(n)}{n}=\sum_{d\mid n}\frac{1}{d} ,\forall n\in \mathbb N^*$$

It’s from a number theory textbook , i’ve tried to use to formula of $\sigma(n)$, which is :

$$\prod_{i=1}^{n} \frac{p_i^{a_i+1}-1}{p_i-1} \text{ Where } n=\prod_{i=1}^{n}p_i^{a_i}$$

But it failed.

Answer 1

The definition of $\sigma(n)$ is, that it is defined as the sum over all divisors of $n$

$$\sigma(n)=\sum_{d\mid n} d$$ But thats the same as saying

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

Dividing by $n$ we get

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

Edit: The equation $$\sigma(n)=\sum_{d\mid n}\frac{n}{d}$$ does hold, because for every divisor $k$ of $n$, there is another divisor $d$ such that $k\cdot d = n$

Therefore, if we sum over the divisors $k$ we can instead sum over $\frac{n}{d}$

Answer 2

An example

Choose $n=6$. On the left hand side we get $\frac{12}{6}=2.$ Now how to we evaluate the right hand side? First write the term out $$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{6}{6}+\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{1+2+3+4+5+6}{6}=\frac{12}{6}=2.$$

Do the same thing but in general.

The general case

We start with the sum. Define $d_j$ as a divisor of $n$.
$$\frac{1}{d_1}+\frac{1}{d_2}+\cdots+\frac{1}{d_{{m-1}}}+\frac{1}{d_{m}}.$$

We know that $d_1=1$ and $d_m=n$, where $d_m$ is the biggest divisor of $n$. Now Rewrite every single term like we did in our example. $$\frac{d_m}{d_m}+\frac{d_{m-1}}{d_m}+\cdots+\frac{d_2}{d_{{m}}}+\frac{d_1}{d_{m}}.$$ Now you may add the terms on in the numerator to get $$\frac{d_m+d_{m-1}+\cdots+d_2+d_1}{n}=\frac{\sigma(d_m)}{d_m}=\frac{\sigma(n)}{n}.$$ And this is your desired result.

Answer 3

The map $d\mapsto n/d$ is a bijection of the set of divisors of $n$ onto itself, hence $$\sigma(n)=\sum_{d\mid n}d=\sum_{d\mid n}\frac{n}{d}=n\sum_{d\mid n}\frac1d.$$