prove that $ \frac{\sigma(1)}{1}+\frac{\sigma(2)}{2}+\dots+\frac{\sigma(n)}{n} \leq 2 n $

Question

[HMMT 2004] For every positive integer $n$, prove that $ \frac{\sigma(1)}{1}+\frac{\sigma(2)}{2}+\dots+\frac{\sigma(n)}{n} \leq 2 n $

If $d$ is a divisor of $i,$ then so is $\frac{i}{d},$ and $\frac{i / d}{i}=\frac{1}{d} .$

Summing over all divisors $d$ of $i$ (which is $\sigma(i)$ ), we see that $\frac{\sigma(i)}{i}$ is the sum of all the reciprocals of the divisors of $i ;$

that is, $ \frac{\sigma(i)}{i}=\sum_{d | i} \frac{1}{d} $ .....

But how they concluded that $ \frac{\sigma(i)}{i}=\sum_{d | i} \frac{1}{d} $ ??? ***

i am getting trouble in understanding this question in recent few days, what i till now understand is that

if $d$ is divisor of $i$ then $i/d$ is also divisor of $i$ , so

$\frac{i / d}{i}=\frac{1}{d} .$ but they are not all divisors of i so how we get *** ,i also tried to take some examples but still can't get it,

can anyone explain this ..

thankyou

Answer 1

Why do you say that “they are not all divisors of $i$”? You have$$\sigma(i)=\sum_{d\mid i}d$$and therefore$$\frac{\sigma(i)}i=\sum_{d\mid i}\frac di.$$But the numbers of the form $\frac di$, with $d\mid i$ are precisely the numbers of the form $\frac1{d'}$, with $d'\mid i$. That is, the are the inverses of the divisors of $i$.