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

Is the next asymptotic term known? That is, is there a monotonic increasing function $f(x)$ such that $$ \sum_{k=1}^n\sigma(k)=\frac{\pi^2}{12}n^2+f(n)+o(f(n)) $$ ? (I would guess something like $f(x)=cx$ if $f$ exists.) Alternately, are there monotonic increasing functions $f(x),g(x)$ with $$ \sum_{k=1}^n\sigma(k)=\frac{\pi^2}{12}n^2+f(n)+\Omega_\pm(g(n)) $$ ?

Apologies for what may be a basic reference question; I've misplaced by copy of Hardy & Wright and lack more advanced references like Apostol.

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What you give is Theorem 324 on page 266 of H+W. They finally prove Gronwall's Theorem in section 22.9, pages 353-354, which is a lim sup result, precursor to Robin's proof that a certain condition on the maximum size of $\sigma(n)$ is equivalent to RH. But they do not seem to typically give more than a principal term and a single error estimate. –  Will Jagy May 17 '12 at 2:32
    
@WillJagy: Thanks for looking that up. I found a copy of Apostol's Introduction to Analytic Number Theory and his Theorem 2.4 is the same as Theorem 324. –  Charles May 17 '12 at 17:55
    
I'm not sure it will help, but Bach and Shallit collect together a bunch of published estimates from a variety of sources, kind of a glossary in the middle of the book, amazon.com/Algorithmic-Number-Theory-Vol-Foundations/dp/… –  Will Jagy May 18 '12 at 0:41

1 Answer 1

up vote 3 down vote accepted

In Apostol's Introduction to Analytic Number Theory he uses $\sum_{q<x}q = x^2/2+O(x)$ to prove this bound as Theorem 3.4. We may be able to do better by using the exact expression $$ \begin{align} \sum_{q<x} q &= \frac{1}{2}\left(\lfloor x \rfloor^2+\lfloor x \rfloor \right) \\ & = \frac{1}{2}\left((x-\{x\})^2+x-\{x\}\right) \\ & = \frac{x^2}{2} - x\{x\}+\frac{x}{2} + O(1) \end{align} $$ where $\{x\}$ is the fractional part of $x$.

Then proceeding in the same way $$ \begin{align} S(n) = \sum_{k\le n}\sigma(k) & = \sum_{d\le n}\sum_{q\le n/d}q \\ & = \sum_{d\le n}\left(\frac{n^2}{2d^2}+\frac{n}{2d}-\frac{n\{n/d\}}{d}+O(1)\right) \\ & = \frac{1}{2}n^2\zeta(2)+\frac{1}{2}n\log(n)-n\sum_{q\le n}\frac{\{n/d\}}{d}+O(n) \end{align} $$

Write $t(n)=n\sum\{n/d\}/d$ for the third term. Since $|\{n/d\}|<1$, $t(n)=O(n\log(n))$.

Let $S(n)=n^2\pi^2/12+F(n)$. Gronwall's Theorem gives us that there are arbitrarily large $N$ with $\sigma(N)>N\log\log(N)$. For such a $N$ we have $$ \begin{align} S(N)-S(N-1) & =\sigma(N)>N\log\log(N) \\ F(N)-F(N-1) & > N \log\log(N)-(2N-1)\pi^2/12 \\ \max(|F(N)|,|F(N-1)|) & > N\log\log(N)/2 - N \end{align} $$ So $F(n)=\Omega(n\log\log(n))$. But $F(n) = n\log(n)/2-t(n)+O(n)$.

Write $t(n)=An\log(n)-f(n)$ with $f(n)=o(n\log(n))$. Then either:

$A\neq \frac{1}{2}$ and $F(n) = \left(\frac{1}{2}-A\right)n\log(n)+o(n\log(n))$

or

$A = \frac{1}{2}$ and $F(n) = f(n) + O(n)$, with $f(n)=o(n\log(n))$ and $f(n)=\Omega(n\log\log(n))$

Unfortunately I can't resolve which case applies, and in fact, if it's the second then I haven't really answered the question after all.

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It is the second case, as it happens. –  Charles May 23 '12 at 14:05
    
That was my guess from calculating some numbers. Do you have a clean answer to your question? I'd be curious to see it now that I have spent some time thinking about this. I was too hasty posting this; I realized I also missed the possibility that it might not be possible to write $t(n)=An\log(n)+o(n\log(n))$. –  Zander May 23 '12 at 23:42
    
I don't have an answer for my question, no. I (now) know that the residual is between $n(log n)^{3/4}$ and $n\log\log n$ but not what the next term might be, if indeed there is a nonoscillating next term. But your answer was helpful regardless, so +1. –  Charles May 24 '12 at 13:18

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