TITLE: Find the sum of reciprocals of divisors given the sum of divisors
QUESTION [12 upvotes]: Let $d_1, d_2, \cdots d_k$ be all the factors of a positive integer '$n$' including $1$ and $n$. Suppose $d_1 + d_2 + d_3+\cdots+d_k = 72$. Then find the value of $\frac{1}{d_1}+\frac{1}{d_2}+\cdots + \frac{1}{d_k}$.
Attempt:
Consider, $d_1, d_2, \cdots d_k$ are the factors of $n$ ascending order. Then
$d_1d_k=n$, $d_2d_{k-1}=n$, $\cdots$, $d_kd_{1}=n$    ............... (1)
Assume, $\frac{1}{d_1}+\frac{1}{d_2}+\cdots + \frac{1}{d_k}=P$
Then $\frac{n}{d_1}+\frac{n}{d_2}+\cdots + \frac{n}{d_k}=nP$ 
i.e 
$d_k+d_{k-1}+\cdots +d_2+d_1=nP$  (using 1)
i.e $72=nP$, or $P=72/n$.
Problem: I have considered that number of divisors of $n$ are even as every divisor $d$ of $n$ has a twin divisor $n/d$. But it is not true if $n$ is a perfect square i.e $n=d^2$ as in this case $n$ has odd number of divisors (for example, $4=2^2$, $4$ has three divisors $1,2 ~\&~ 4$).

REPLY [3 votes]: Generally the sum of the reciprocals of the divisors of $n$ is equal to $\frac{\sigma(n)}{n}$ where $\sigma$ is the sum of divisors function.  This quantity is sometimes referred to as the abundancy ratio or abundancy index of $n$.  It can be used to tell whether $n$ is abundant, deficient, or perfect.