The average order of $\sigma(n)$.

Question

I'm reading through Apostol's wonderful book "Introduction to Analytic Number Theory" but became confused when reading the statement of theorem 3.4. It states that $$ \sum_{n\leq x}\sigma(n) = \frac{1}{2}\zeta(2)x^2 + O(x\log(x)). $$

This would imply that the average order of $\sigma(n)$ is $\frac{1}{2}\zeta(2)n = \frac{\pi^2}{12}n < n$. How can the average order of $\sigma(n)$ be less than $n$, when $\sigma(n) > n$ for all $n$?

Wikipedia claims that the average order of $\sigma(n)$ is $\zeta(2)n$ under "Examples" here

but further down the page, they claim that for any $\alpha > 0$ we have that $$ \sum_{n\leq x}\sigma_\alpha(n) = \frac{\zeta(\alpha+1)}{\alpha+1}x^{\alpha+1} + O(x^\beta) $$ where $\beta = \max(1,\alpha)$ which gives the same result as Apostol for $\alpha = 1$. What am I not understanding?

Answer 1

Indeed you have $\sigma(n)>n$ for all $n$, but for 'most' values of $n\leq x$ you have $\sigma(n)<x$. So it should come as no surprise that the average value of $\sigma(n)$ for all $n\leq x$ is less than $x$.

Answer 2

The result in Apostol is correct.

The average is over all the integers up to $x$, so we should expect it to be smaller than $x$.