1
$\begingroup$

I was wondering about studying number theory. Take a function $d$ as follow:

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

And I will consider the limit of this function.

$ \lim_{n \to \infty} d(n) $

I expect this value to be $\sum \frac{1}{n} = \infty$

What about my thought? If it is true, please explain why..!

  • 0
    If $n=\prod_{r=1}^mp_r^{k_r}$ $$\sum_{d|n}\dfrac1d=\dfrac1n\sum_{d|n}d=\prod_{r=1}^m\dfrac{p_r^{k_r+1}-1}{(p_r-1)p_r^{k_r}}$$ – 2017-01-17
  • 4
    If $n=p$ is prime then $d(p)=1+\frac 1p$. Thus there are infinitely many $n$ for which $d(n)$ is near $1$. – 2017-01-17
  • 2
    Your function is essentially $\displaystyle{\sigma_1(n)\over n}$. There is no limit; as for the _upper limit_, it is $\infty$ indeed. – 2017-01-17

3 Answers 3

0

The other answers make this way harder than it needs to be. Just note that $d(n!)\geq\sum_{k=1}^n\frac{1}{k}$, since $k\mid n!$ for $k=1,\dots,n$. Since the infinite sum $\sum_{k=1}^\infty\frac{1}{k}$ diverges to $\infty$, $d(n!)$ goes to $\infty$ as $n$ goes to $\infty$. In particular, $$\limsup_{n\to\infty}d(n)=\infty.$$ On the other hand, $d(n)\geq 1$ for all $n$, and $d(n)=1+\frac{1}{n}$ if $n$ is prime. Since $1+\frac{1}{n}$ converges to $1$ as $n$ gets large and there are infinitely many primes, we get $$\liminf_{n\to\infty}d(n)=1.$$

So it is not correct that $\lim\limits_{n\to\infty}d(n)=\infty$; that is the lim sup, but the limit itself has no value since the lim inf is different.

1

Due to a result of Euler $\sum_{p\text{ prime}}\frac{1}{p}=\infty$. So for $1\neq N\in\mathbb{N}$ there exists $r\in\mathbb{N}$ such that if $p_1,...,p_r$ are the first $r$ primes and $n=p_1...p_r$ then

$N<\frac{1}{p_1}+...+\frac{1}{p_r}

but $d(m)\leq N$ for some $m>n$.

So $\lim \sup_{n\rightarrow \infty} d(n)=\infty$. And of course $\lim\inf_{n\rightarrow\infty}d(n)=1$ as explained by lulu.

1

Our function is: $$\sum_{d\mid n}\frac{1}{d} = \sum_{d\mid n}\frac{d}{n} = \frac{1}{n}\sum_{d\mid n}d = \frac{\sigma_1(n)}{n} $$ so a multiplicative function. If we restrict such function to square-free numbers by assuming $n = p_1 p_2\cdots p_k$ with $p_1,p_2,\ldots,p_k$ being distinct primes, we get: $$ \frac{\sigma_1(n)}{n} = \prod_{p\mid n}\left(1+\frac{1}{p}\right)=\prod_{p\mid n}\left(1-\frac{1}{p}\right)^{-1}\prod_{p\mid n}\left(1-\frac{1}{p^2}\right)$$ where $\prod_{p\in\mathcal{P}}\left(1-\frac{1}{p^2}\right)$ is a convergent series, equal to $\frac{1}{\zeta(2)}=\frac{6}{\pi^2}$. In particular, $$ \frac{\sigma_1(n)}{n}\geq\frac{6}{\pi^2}\prod_{p\mid n}\left(1-\frac{1}{p}\right)^{-1} $$ and if we assume that $p_1,\ldots,p_k$ are the primes up to $x$, $$ \frac{\sigma_1(n)}{n}\geq \frac{6}{\pi^2}\sum_{n\leq x}\frac{1}{n}\geq \frac{6}{\pi^2}\log(x).$$ That proves our function is unbounded, we do not need anything too sophisticated.
On the other hand, $$ \liminf_{n\to +\infty}\frac{\sigma_1(n)}{n} = 1 $$ follows from considering $n=p^m$.