3
$\begingroup$

The wolframalpha page on the divisor function, http://mathworld.wolfram.com/DivisorFunction.html

Lists the asymptotic estimate,

$$\sum_{n\leq x}\sigma(n)=\frac{\pi^2}{12}x^2+O(x\ln(x))$$

However, I have tried to get the same estimate on my own by interchanging the summation on the left hand side and using the dirichlet hyperbola method.

Doing this the best estimate I can seem to get is,

$$\sum_{n\leq x}\sigma(n)=\frac{\pi^2}{12}x^2+O(x^{3/2})$$

Were non 'elementary' techniques used to obtain the first asymptotic? Or is it a typo?

$\endgroup$
4
$\begingroup$

It's Theorem $324$ in Hardy/Wright, using what I guess is the hyperbola method. The proof is essentially:

$$\begin{align} \sum_{l=1}^n \sigma(l) &= \sum_{x=1}^n \sum_{y \leqslant n/x} y = \sum_{x=1}^n \frac12 \left\lfloor\frac{n}{x}\right\rfloor \left(\left\lfloor\frac{n}{x}\right\rfloor+1\right)\\ &= \frac12 \sum_{x=1}^n \left(\frac{n}{x} + O(1)\right)\left(\frac{n}{x} + O(1)\right) = \frac12 n^2 \sum_{x=1}^n \frac{1}{x^2} + O\left(n\sum_{x=1}^n \frac{1}{x}\right) + O(n). \end{align}$$

Now $\displaystyle \sum_{x=1}^n \frac{1}{x^2} = \sum_{x=1}^\infty \frac{1}{x^2} + O\left(\frac{1}{n}\right) = \frac16 \pi^2 + O\left(\frac{1}{n}\right)$,
by $(17.2.2)$, and $\displaystyle \sum_{x=1}^n \frac{1}{x} = O(\log n)$.

Hence $\sum_{l=1}^n \sigma(l) = \frac{1}{12}\pi^2n^2 + O(n\log n)$.

$\endgroup$
  • $\begingroup$ I see my error now, they didn't use the hyperbola method at all. I guess using it only made the estimate worse lol $\endgroup$ – Ethan Aug 3 '13 at 22:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.