Upper and Lower bounds of Sum of Sum-of-Divisors 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?
 A: 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 $x^2$
