Sum of the sum-of-divisors function - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T09:16:41Z http://mathoverflow.net/feeds/question/100027 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100027/sum-of-the-sum-of-divisors-function Sum of the sum-of-divisors function Charles 2012-06-19T18:00:33Z 2012-06-24T14:14:27Z <p>I was looking at <a href="http://rd.springer.com/article/10.1007/BF01630684" rel="nofollow">the abstract</a> of a paper [1] which claims that [2] and [3] prove $$\sum_{n\le x}\sigma(n)-\frac{\pi^2}{12}x^2=\Omega(x\log\log x).$$</p> <p>But I cannot find the above&mdash;or indeed, anything approaching it&mdash;in [2]. Have I missed something?</p> <p>The paper [3] clearly discusses the appropriate function and presumably gives the indicated result. I must decipher its notation, though: the author seems to use $\sigma(n)$ to denote what would usually be written $\sigma_{-1}(n)=\sigma(n)/n.$</p> <h2>References</h2> <p>[1] Y.-F. S. Pétermann, "An Ω-theorem for an error term related to the sum-of-divisors function", <em>Monatshefte für Mathematik</em> 103:2 (1987), pp. 145-157.</p> <p>[2] T. H. Gronwall, "Some asymptotic expressions in the theory of numbers", <em>Trans. Amer. Math. Soc.</em> <strong>14</strong> (1913), pp. 113–122. <a href="http://www.jstor.org/stable/1988773" rel="nofollow">JSTOR</a></p> <p>[3] S. Wigert, <a href="http://home.us.archive.org/stream/actamathematica37upps#page/n145/mode/2up" rel="nofollow">Sur quelques fonctions arithmétiques</a>, <em>Acta Math.</em> <strong>37</strong> (1914), pp. 113–140.</p> http://mathoverflow.net/questions/100027/sum-of-the-sum-of-divisors-function/100044#100044 Answer by Igor Rivin for Sum of the sum-of-divisors function Igor Rivin 2012-06-19T20:09:30Z 2012-06-19T20:09:30Z <p>This is proved in G. Tenenaum's book (introduction to analytic and probabilistic number theory), page 39 (section 3.3, theorem 3). I agree that Gronwall's paper, other than the fact that it studies the same function, seem to be completely unrelated.</p> http://mathoverflow.net/questions/100027/sum-of-the-sum-of-divisors-function/100071#100071 Answer by Gerry Myerson for Sum of the sum-of-divisors function Gerry Myerson 2012-06-20T04:26:28Z 2012-06-20T04:26:28Z <p>It is not clear whether you are asking for a proof/reference for the displayed formula, or an evaluation of the contents of the cited papers. </p> <p>Dickson's History, Volume 1, page 323, says Wigert proved $$\sum_{n\le x}\sigma(n)={\pi^2x^2\over12}+x((1/2)\log x-\psi(x))+O(x)$$ where $$\psi(x)=x\sum_{n\gt x}{1\over n^2}+\sum_{n\le x}{1\over n}\rho\left({x\over n}\right)$$ and $\rho(x)$ is the fractional part of $x$. Further, for $x$ sufficiently large, $$((1/4)-\epsilon)\log x\lt\psi(x)\lt((3/4)+\epsilon)\log x$$ It seems to me that this gives a poorer error term than the one in your display. Dickson also says Landau gave corrections and simplifications to Wigert's proofs, Gottingsche gelehrte Anzeigen 177 (1915) 377-414. </p> http://mathoverflow.net/questions/100027/sum-of-the-sum-of-divisors-function/100127#100127 Answer by quid for Sum of the sum-of-divisors function quid 2012-06-20T12:23:51Z 2012-06-20T12:23:51Z <p>The clue to understanding the relevance of the quoted results seems to be given in Remark 2 of Pétermann's paper (at the very end). Where it is said that a result on the limes superior of <code>$\sigma_{-1}(x)/ \log \log x$</code> implies an Omega-result on the error term <code>$E_{-1}$</code> . (And thus <code>$E_{1}$</code> which is the one in the question.) This result mentioned in Remark 2, also actually appears in the paper of Gronawall, see Eq. (25) there; except it is staed for <code>$\sigma_1$</code> , but this translates directly as commented at the beginning of that paper where the relation between <code>$\sigma_{a}$</code> and <code>$\sigma_{-a}$</code> is mentioned. </p> <p>ps. This was written a bit quickly, I hope I still got the details right, but in any case this Remark 2 seems to be helpful in understanding the relation to the earlier papers.</p>