What is the sum of the reciprocal of all of the factors of a number? Suppose I have some operation $f(n)$ that is given as

$$f(n)=\sum_{k\ge1}\frac1{a_k}$$

Where $a_k$ is the $k$th factor of $n$.
For example, $f(100)=\frac11+\frac12+\frac14+\frac15+\frac1{10}+\frac1{20}+\frac1{25}+\frac1{50}+\frac1{100}=\frac{217}{100}$
$f(101)=\frac11+\frac1{101}=\frac{102}{101}$
$f(102)=\frac11+\frac12+\frac13+\frac16+\frac1{17}+\frac1{34}+\frac1{51}+\frac1{102}=\frac{216}{102}$
I was wondering if it were possible to plot a graph of $f(n)$ and wondered if there were any interesting patterns.  I was also wondering if there is a closed form representation and if $\lim_{n\to\infty}f(n)$ could be evaluated or determined to be finite or not or any other interesting things that might happen in this limit.
Secondly, I was wondering about another similar series, which considers $b_k$ as the $k$th prime factor of $n$.

$$p(n)=\sum_{k\ge1}\frac1{b_k}$$

What can we determine about this series?
 A: Let $X_k$ be the product of the first $k$ primes.  Let $Z_k$ be the sum of the reciprocals of the first $k$ primes.  Then clearly $f(X_k)>Z_k$, and it's well known that $Z_k$ is unbounded, so $f(a_k)$ cannot have a finite limit.  On the other hand, if $P_k$ is the $k$'th prime, then $f(P_k)$ clearly goes to $1$.  Therefore $f(a_k)$ cannot have a limit other than $1$.  Therefore $lim_{k\rightarrow\infty}a_k$ cannot exist.
A: Just for the fun of it, I've graphed this up to $n=18$
Here is the data:

As you can see, the points move all over the place. Over this domain, it does seem to have a local maximum at every even number though. Perhaps this is because they have the advantage of a plus $1/2$
Here is the table I made if anyone wants to double check it:

Also, here is the link to the graph
A: Here is the plot of $f(n)=\frac{\sigma(n)}{n}$ .
Patterns in these plots are amazing!
For $1000$  
 
And here is for $100,000$

This diagram shows the $\lim_{x\to \infty} f_n$ isn't plausible !
A: Note that $n\cdot f(n)$ is the sum of the factors of $n$ (written in a different order), which is denoted by $\sigma(n)$. Thus, $\displaystyle f(n)={\sigma (n)\over n}$.
