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?

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, $\displaystyle f(n)={\sigma (n)\over n}$.

Ramanujan included this in his original paper on Highly Composite Numbers, originally 1915. http://math.univ-lyon1.fr/~nicolas/ramanujanNR.pdf However, this was in a section left out because of paper shortages.

Let's see, I asked about this on MO https://mathoverflow.net/questions/137865/estimate-term-in-ramanujan-lost-notebook-classic-analytic-number-theory but did not quite get what I wanted, so I wrote to Nicolas. He's a nice man, but he had never heard of me, and the websites I mentioned were unknown to him. Sigh. Anyway, he did answer.

In brief, Ramanujan's construction allows us to produce a sequence of numbers, each new one the previous one times a prime, so that the function $\sigma(n)/n$ is surprisingly large for $n$ of that size. In turn, this gives explicit bounds on the function.

For numerical experiments of your own, the easiest way to approximate the numbers in this sequence is simply to take $$ n = \operatorname{lcm} \{1,2,3, \ldots, k \} $$ and put $n$ into the sequence when it increases, which happens only when $k$ is a prime or prime power. Extremely approximately, $n \approx e^k.$ From Robin's criterion and related stuff, we will have $$ \frac{\sigma(n)}{n} \approx e^\gamma \log \log n \approx e^\gamma \log k, $$ where $ n = \operatorname{lcm} \{1,2,3, \ldots, k \} .$ Note that $e^\gamma \approx 1.7810724.$ Also note that it is the Prime Number Theorem that says that $\log n \approx k.$

Did it myself:

2 n = 2 =  2  function:  1.5 over log k:  2.16404
3 n = 6 =  2 3  function:  2 over log k:  1.82048
4 n = 12 =  2^2 3  function:  2.33333 over log k:  1.68314
5 n = 60 =  2^2 3 5  function:  2.8 over log k:  1.73974
7 n = 420 =  2^2 3 5 7  function:  3.2 over log k:  1.64447
8 n = 840 =  2^3 3 5 7  function:  3.42857 over log k:  1.64879
9 n = 2520 =  2^3 3^2 5 7  function:  3.71429 over log k:  1.69044
11 n = 27720 =  2^3 3^2 5 7 11  function:  4.05195 over log k:  1.68979
13 n = 360360 =  2^3 3^2 5 7 11 13  function:  4.36364 over log k:  1.70126
16 n = 720720 =  2^4 3^2 5 7 11 13  function:  4.50909 over log k:  1.62631
17 n = 12252240 =  2^4 3^2 5 7 11 13 17  function:  4.77433 over log k:  1.68513
19 n = 232792560 =  2^4 3^2 5 7 11 13 17 19  function:  5.02561 over log k:  1.70681
23 n = 5354228880 =  2^4 3^2 5 7 11 13 17 19 23  function:  5.24412 over log k:  1.6725
25 n = 26771144400 =  2^4 3^2 5^2 7 11 13 17 19 23  function:  5.41892 over log k:  1.68348
27 n = 80313433200 =  2^4 3^3 5^2 7 11 13 17 19 23  function:  5.55787 over log k:  1.68633
29 n = 2329089562800 =  2^4 3^3 5^2 7 11 13 17 19 23 29  function:  5.74952 over log k:  1.70746
31 n = 72201776446800 =  2^4 3^3 5^2 7 11 13 17 19 23 29 31  function:  5.93499 over log k:  1.72831
32 n = 144403552893600 =  2^5 3^3 5^2 7 11 13 17 19 23 29 31  function:  6.03071 over log k:  1.7401
37 n = 5342931457063200 =  2^5 3^3 5^2 7 11 13 17 19 23 29 31 37  function:  6.1937 over log k:  1.71527
41 n = 219060189739591200 =  2^5 3^3 5^2 7 11 13 17 19 23 29 31 37 41  function:  6.34477 over log k:  1.70854
43 n = 9419588158802421600 =  2^5 3^3 5^2 7 11 13 17 19 23 29 31 37 41 43  function:  6.49232 over log k:  1.72613
47 n = 442720643463713815200 =  2^5 3^3 5^2 7 11 13 17 19 23 29 31 37 41 43 47  function:  6.63046 over log k:  1.72213
49 n = 3099044504245996706400 =  2^5 3^3 5^2 7^2 11 13 17 19 23 29 31 37 41 43 47  function:  6.74886 over log k:  1.73411
53 n = 164249358725037825439200 =  2^5 3^3 5^2 7^2 11 13 17 19 23 29 31 37 41 43 47 53  function:  6.8762 over log k:  1.73191
59 n = 9690712164777231700912800 =  2^5 3^3 5^2 7^2 11 13 17 19 23 29 31 37 41 43 47 53 59  function:  6.99274 over log k:  1.71494

In comparison, the function for, say, $n$ prime is very small, just $1 + (1/n).$

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.