I want to know an upper bound of this function: $$\sum_{n\le x}\sigma(n)$$ where $\sigma(x)$ is the sum of divisors.

Are there any known upper and lower bounds?

  • $\begingroup$ I'm guessing it's an asymptotic, meaning the ratio between them converges to $1$ as $x\to\infty$. That means it may very well be neither an upper or lower bound, even eventually - sometimes estimates can oscillate above and below the true value infinitely often. (I have no idea if that's the case here, it could actually be a bound too.) $\endgroup$ – runway44 Jun 27 at 2:30
  • $\begingroup$ @runway44 Ok. Thank you. $\endgroup$ – Quote Dave Jun 27 at 2:31

This sum actually has a surprisingly nice upper and lower bound. It turns out

$$\frac{x(x+1)}{2} \le \sum_{n\le x}\sigma(n) \le x^2$$

You can verify this for yourself with a python script such as this one

Now let's discuss how I arrived at these bounds:

We can reframe this problem to make it a little simpler to think about. Instead of adding up $\sigma(i)$ for every $i$ from $1$ to $x$, we can instead try to find out how many times each factor(from $1$ to $x$) is counted, and use this to find our answer.

For example in the case of $x=9$ we know:

$1$ will be counted $9$ times (at $1,2,3,4,5,6,7,8,9$)

$2$ will be counted $4$ times (at $2,4,6,8$)

$3$ will be counted $3$ times (at $3,6,9$)

$4$ will be counted $2$ times (at $4,8$)


In general each factor $k$ is counted $\lfloor \frac{x}{k} \rfloor$ times.

This means we can rewrite our sum as

$$\sum_{k=1}^x k \cdot \lfloor \frac{x}{k} \rfloor$$

We can set $ \lfloor \frac{x}{k} \rfloor=1$ to get our lower bound of $\frac{x(x+1)}{2}$

And we can set $ \lfloor \frac{x}{k} \rfloor= \frac{x}{k}$ to get our upper bound of $x^2$

  • $\begingroup$ Wow, thank you! $\endgroup$ – Quote Dave Jul 1 at 2:33
  • $\begingroup$ No problem the question was pretty fun to figure out. $\endgroup$ – Anirudh Jul 1 at 3:20
  • $\begingroup$ However $x^2$ is not the tightest bound. $\endgroup$ – Quote Dave Jul 5 at 16:44
  • $\begingroup$ Is there a better one? $\endgroup$ – Anirudh Jul 5 at 19:21

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