Let $d(n)$ be the number of divisors of an integer $n$.
Does there exists a bound for $\sum_{k\leq n}d^2(k)$?
I saw in a paper of Barry and Louboutin that the asymptotics is $\frac1{\pi^2}nlog^3n$ but there was no proof nor a reference or any indication that this is the best possible bound.
Any idea where can I find this?
Thanks.
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2 Answers
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This can be proved by Perron's formula, making use of the properties of the associated Dirichlet series $$ g(s):=\sum_{k=1}^\infty\frac{d^2(k)}{k^s},\quad\Re s>1. $$ We have an Euler product decomposition over the primes $p$, $$ g(s)=\prod_p\left(\sum_{m=0}^\infty\frac{d^2(p^m)}{p^{ms}}\right)=\prod_p\left(\sum_{m=0}^\infty\frac{(m+1)^2}{p^{ms}}\right)=\prod_p\frac{1-p^{-2s}}{(1-p^{-s})^4},\quad \Re s>1.$$ This shows that $g(s)$ has meromorphic extension to $\mathbb{C}$, namely $$ g(s)=\frac{\zeta(s)^4}{\zeta(2s)}. $$ In particular, $(s-1)^4g(s)$ is holomorphic and bounded in any half-plane $\Re s>\sigma_0>1/2$, and at $s=1$ it takes the value $1/\zeta(2)$. Combining Perron's formula, Cauchy's theorem, the residue theorem, and a contour shift, it is not hard to deduce that $$ \sum_{k=1}^nd^2(k)\sim\frac{1/\zeta(2)}{3!}n(\log n)^3=\frac{1}{\pi^2}n(\log n)^3.$$ The $3!$ and the $(\log n)^3$ come from the third term in the Taylor expansion $$ n^{s-1}=\sum_{r=0}^\infty \frac{(\log n)^r}{r!}(s-1)^r.$$
Of course the same method yields secondary terms with smaller powers of $\log n$. Finding a good error term next to these is a difficult task and goes under the label "Dirichlet's divisor problem". You can learn about these things in textbooks of analytic number theory (e.g. Tenenbaum or Montgomery-Vaughan). Hope this helps.
answered Mar 11, 2014 at 20:23
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1$\begingroup$ We can get an estimate for $\sum d(k)$ using only methods from the first couple of weeks of an intro undergrad elementary Number Theory class. I wonder whether there isn't a similar method for $\sum d^2(k)$. $\endgroup$ Commented Mar 11, 2014 at 22:24
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1$\begingroup$ @Gerry: We can certainly derive the right order of magnitude by more elementary techniques. On the other hand, I believe the constant $\pi^{-2}$ is subtle. That is, getting this constant is much the same as proving $\zeta(2)=\pi^2/6$. $\endgroup$ Commented Mar 11, 2014 at 23:16
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1$\begingroup$ Thanks. $\zeta(2)$ certainly won't be evaluated in the first couple of weeks of intro Number Theory, but it can be evaluated without the full panoply of "Perron's formula, Cauchy's theorem, the residue theorem, and a contour shift." $\endgroup$ Commented Mar 12, 2014 at 2:33
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2$\begingroup$ In fact, the proof given above can be greatly generalized to give the asymptotics of $\sum_{n\le x}f(n)$, if $f$ is multiplicative and has a given average value $v$ on primes. (Here $f=d^2$ and $v=4$.) This is the context of the Selberg-Delange method (see Tenenbaum's book "intro to analytic and probabilistic number theory"). $\endgroup$ Commented Mar 12, 2014 at 3:01
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1$\begingroup$ @Gerry: I agree. Still, I don't think an elementary proof would be as straightforward as the first moment. If you can see an easy approach, assuming the value of $\zeta(2)$, please give it as a second answer. Also note that Cauchy's formula etc. are part of the basic curriculum (introductory complex analysis). $\endgroup$ Commented Mar 12, 2014 at 7:47
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Inspired by Gerry Myerson's comment, I present below an elementary argument that does not use any complex analysis.
Our starting point is the identity $$ d^2(k)=\sum_{k=l^2m}\mu(l)d_4(m), $$ where $d_4(m)$ enumerates the number of decompositions $m=m_1m_2m_3m_4$. By the multiplicative nature of this equation, it suffices to verify it when $m=p^r$ is a prime power. In this case the identity boils down to $$ (r+1)^2=\binom{r+3}{3}-\binom{r+1}{3}, $$ which is straightforward to check. With the above identity at hand, we rewrite $$ \sum_{k\leq n}d^2(k)=\sum_{l^2m\leq n}\mu(l)d_4(m)=\sum_{l\leq\sqrt{n}}\mu(l)\sum_{m\leq\frac{n}{l^2}}d_4(m). $$ We approximate the inner sum by Dirichlet's hyperbola method (cf. Exercise 2 in Section 1.5 of Iwaniec-Kowalski: Analytic Number Theory), and we obtain $$ \sum_{k\leq n}d^2(k)=\sum_{l\leq\sqrt{n}}\mu(l)\frac{n}{6l^2}\left\{\log^3\left(\frac{n}{l^2}\right)+O\left(\log^2\left(\frac{n}{l^2}\right)\right)\right\}. $$ It follows that $$ \sum_{k\leq n}d^2(k)=\frac{n\log^3(n)}{6}\left(\sum_{l\leq\sqrt{n}}\frac{\mu(l)}{l^2}\right)+O\left(n\log^2(n)\right) =\frac{n\log^3(n)}{6\zeta(2)}+O\left(n\log^2(n)\right).$$ Finally, using $\zeta(2)=\pi^2/6$ as a black box, we obtain the formula asked by the OP.
answered Mar 12, 2014 at 10:36
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$\begingroup$ I vaguely remember reading somewhere that Ramanujan had computed the asymptotics for all moments of the divisor function (though maybe just upper bounds?) This would certainly have been without complex analysis... $\endgroup$ Commented Mar 12, 2014 at 16:24
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1$\begingroup$ @Denis: The above calculation can be generalized. That is, $d^r$ can be expressed as the convolution of $d_{2^r}$ with some explicit arithmetic function $f_r$. This yields the asymptotic $c_r n\log^{2^r-1}(n)$ for the $r$-th moment, where $c_r$ equals $\sum_m f_r(m)/m$ divided by $(2^r-1)!$. $\endgroup$ Commented Mar 12, 2014 at 17:55