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What is known about the asymptotic behavior of $$ -\frac{\pi^2}{18}x^3+\sum_{n\le x}n\sigma(n) ? $$

It seems to be $O(x^{2+\varepsilon})$ but I cannot prove this.

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2 Answers 2

up vote 6 down vote accepted

Look at the corresponding Dirichlet series and make use of Perron's formula to get the asymptotic. $$f(n) = n \sigma(n) \implies F(s) = \sum_{n=1}^{\infty} \dfrac{n \sigma(n)}{n^s} = \sum_{n=1}^{\infty} \dfrac{\sigma(n)}{n^{s-1}} = \zeta(s-1)\zeta(s-2)$$ $F(s)$ has a pole at $s=2$ and $s=3$. Now making use of Perron's formula, the leading order terms of $\displaystyle \sum_{n \leq x} n \sigma(n)$ are the leading order terms in the integral (keeping $c>3$) $$\dfrac1{2 \pi i}\displaystyle \int_{c-i T}^{c+i T} \zeta(s-1) \zeta(s-2) \dfrac{x^s}{s} ds$$ for a sufficiently large and appropriately chosen $T$.

For $c>3$, we have $$\dfrac1{2 \pi i} \int_{c - i T}^{c + i T} \zeta(s-1) \zeta(s-2) \dfrac{x^s}s dx = \sum_{n=1}^{\infty} \dfrac{ n\sigma(n) }{2 \pi i} \int_{c-iT}^{c+iT}\left(\dfrac{x}{n} \right)^s \dfrac{ds}s$$

$$\sum_{n=1}^{\infty} \dfrac{ n\sigma(n) }{2 \pi i} \int_{c-iT}^{c+iT}\left(\dfrac{x}{n} \right)^s \dfrac{ds}s = \sum_{n \leq x} n \sigma(n) + \mathcal{O} \left( \sum_n \left(\left( \dfrac{x}{n}\right)^c \dfrac{n \sigma(n)}{T \log \left(\dfrac{x}{n} \right)} \right) \right)$$

Hence, we have $$\dfrac1{2 \pi i} \int_{c - i T}^{c + i T} \zeta(s-1) \zeta(s-2) \dfrac{x^s}s dx = \sum_{n \leq x} n \sigma(n) + \mathcal{O} \left( \sum_n \left(\left( \dfrac{x}{n}\right)^c \dfrac{n \sigma(n)}{T \log \left(\dfrac{x}{n} \right)} \right) \right)$$

Now we look at the error term. We will use the following bounds for $\zeta$. $$\left \vert \zeta(1+k+it) \right \vert \ll \dfrac1{k}$$ $$\left \vert \zeta(k+it) \right \vert \ll (1+\vert t \vert)^{1-k} \log (1+ \vert t \rvert)$$ where $k > 0$.

First we will look at the error $$ \sum_n \left(\left( \dfrac{x}{n}\right)^c \dfrac{n \sigma(n)}{T \log \left(\dfrac{x}{n} \right)} \right)$$ split this into two pieces. Say $\vert n - x \vert \leq 0.1x$ and $\vert n - x \vert > 0.1x$.

In the case, $\vert n - x \vert > 0.1x$, the error is $$ \sum_{n:\vert n - x \vert > 0.1x} \left(\left( \dfrac{x}{n}\right)^c \dfrac{n \sigma(n)}{T \log \left(\dfrac{x}{n} \right)} \right) \ll \dfrac{x^c}{T} \sum_n \dfrac{\sigma(n)}{n^{c-1}} = \dfrac{x^c}{T} \zeta(c-1) \zeta(c-2)$$

Choosing $c = 3 + \dfrac1{\log x}$.

$$ \sum_{n:\vert n - x \vert > 0.1x} \left(\left( \dfrac{x}{n}\right)^c \dfrac{n \sigma(n)}{T \log \left(\dfrac{x}{n} \right)} \right) \ll \dfrac{x^\left(3 + 1/\log x\right)}{T} \zeta(2) \zeta(1 + 1/ \log(x)) \ll \dfrac{x^3}{T} \log(x)$$

In the case, $\vert n - x \vert \leq 0.1x$, taking $x = \text{integer} + 1/2$, the error is $$ \sum_{n:\vert n - x \vert \leq 0.1x} \left(\left( \dfrac{x}{n}\right)^c \dfrac{n \sigma(n)}{T \log \left(\dfrac{x}{n} \right)} \right) \ll \dfrac{x^c}{T} \sum_{n:\vert n - x \vert \leq 0.1x} \left(\left( \dfrac1{n^{c-2 - \epsilon}}\right) \dfrac1{\log \left(\dfrac{x}{n} \right)} \right) \ll \dfrac{x^3}{T} \log(x)$$

Now we need to evaluate some asymptotic for $\displaystyle \dfrac1{2 \pi i} \int_{c - i T}^{c + i T} \zeta(s-1) \zeta(s-2) \dfrac{x^s}s ds$. Let $d \in (2,3)$, then we have that $$\displaystyle \dfrac1{2 \pi i} \int_{c - i T}^{c + i T} \zeta(s-1) \zeta(s-2) \dfrac{x^s}s ds = \underbrace{\zeta(2) \dfrac{x^3}3}_{\text{Leading order term}}$$ $$+ \underbrace{\displaystyle \dfrac1{2 \pi i} \int_{d - i T}^{d + i T} \zeta(s-1) \zeta(s-2) \dfrac{x^s}s ds}_{\text{Vertical Integral}}$$ $$+ \underbrace{\displaystyle \dfrac1{2 \pi i} \int_{c - i T}^{d - i T} \zeta(s-1) \zeta(s-2) \dfrac{x^s}s ds + \displaystyle \dfrac1{2 \pi i} \int_{c + i T}^{d + i T} \zeta(s-1) \zeta(s-2) \dfrac{x^s}s ds}_{\text{Horizontal Integrals}}$$ Let $d = 2+k$, where $k>0$.

Bounds for the vertical integral:

$$\left \lvert \displaystyle \dfrac1{2 \pi i} \int_{d - i T}^{d + i T} \zeta(s-1) \zeta(s-2) \dfrac{x^s}s ds \right \rvert \leq \dfrac1{2 \pi} \int_{-T}^{T} \vert \zeta(1+k+it) \vert \vert \zeta(k+it) \vert \dfrac{x^{2+k}}{\sqrt{(2+k)^2 + t^2}} dt$$ $$ \ll \dfrac1{2 \pi k}\int_{-T}^{T} \dfrac{(1+\vert t \vert)^{1-k} x^{2+k}}{1 + \vert t \vert} \log(1 + \vert t \vert) dt \ll x^{2+k} T^{1-k} \log (T)$$

Bounds for the horizontal integral:

$$\left \vert \displaystyle \dfrac1{2 \pi i} \int_{c - i T}^{d - i T} \zeta(s-1) \zeta(s-2) \dfrac{x^s}s ds \right \vert \leq \dfrac1{2 \pi} \int_c^d \dfrac{\vert \zeta(\sigma -1 + iT) \vert \vert \zeta(\sigma -2 + iT) \vert x^{\sigma}}{\sqrt{T^2 + \sigma^2}} d \sigma \ll x^{2+k} T^{1-k} \log (T) + \dfrac{x^3 \log(x)}{T}$$

Hence, now we have $$\sum_{n \leq x} n \sigma(n) = \dfrac{\pi^2}{18}x^3 + \mathcal{O} \left(x^{2+k} T^{1-k} \log (T) + \dfrac{x^3 \log(x)}{T}\right)$$ Now the optimal choice for $k$ and $T$ is $\epsilon>0$ and $T = \sqrt{x}$ respectively. This gives us an error of $\mathcal{O}(x^{2.5 + \epsilon})$.

Hence, you have that $$\sum_{n \leq x} n \sigma(n) = \dfrac{\pi^2}{18}x^3 + \mathcal{O} \left(x^{2.5+\epsilon}\right)$$

You can also try to use elementary methods using partial summation and hyperbola method but I believe it is a bit too much work and using Perron's formula is simpler.

EDIT Deleted some incorrect stuff which I wrote initially.

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How can we keep $c>3$ and simultaneously shift it left of $s=2$? –  Zander Jun 19 '12 at 18:55
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@Zander As an example, to evaluate $\displaystyle \dfrac1{2 \pi i}\int_{(c)} \dfrac{y^s}s ds$ for $y>1$ and $c>0$, you shift $c$ to the left of $0$ and move it towards $-\infty$, in which case the integrand becomes very close to zero but in the process you pick up the pole at $s=0$ (i.e. evaluate the appropriate residue at $s=0$), which gives you that $\displaystyle \dfrac1{2 \pi i}\int_{(c)} \dfrac{y^s}s ds = 1$ for $y>1$ and $c>0$.You may want to look up Mellin transform (en.wikipedia.org/wiki/Mellin_transform) & Perron's formula (en.wikipedia.org/wiki/Perron%27s_formula). –  user17762 Jun 19 '12 at 19:09
    
@Zander When you shift, from $c>3$ to $c<2$, you then pick up the poles from the integrand. i.e. $$\displaystyle \dfrac1{2 \pi i}\int_{(c>3)} F(s) ds = \left( \text{Residue of } F(s) \text{ at s= 3}\right) + \left(\text{Residue of } F(s) \text{ at s= 2} \right)+ \displaystyle \dfrac1{2 \pi i}\int_{(c<2)} \dfrac{y^s}s ds$$ where $(c)$ denotes the $c$ line $c-i \infty$ to $c + i\infty$. And sorry for the multiple pings :) –  user17762 Jun 19 '12 at 19:15
    
@EricNaslund Ok. So I can move my integral only to $c \in (2,3)$ since if I move the integral to the left of $2$, I need bounds for $\zeta$ to the left of the $0$ line which will not behave nicely to give us and error of $o(x^2)$? Is this right? –  user17762 Jun 19 '12 at 19:25
    
@EricNaslund I think I saw your comment. I wanted to know if what I have written is actually correct. Could you let me know if the asymptotic is incorrect? Else I will try to work it out and modify it later. –  user17762 Jun 19 '12 at 19:29

See my blog post regarding the average of $\sigma(n)$. This post is a two part series, part I looks at the upper bound, and Part II proves Pétermann's lower bound, which is significantly more difficult. All results regarding $n\sigma(n)$ follow right away from partial summation.

In Part I, the hyperbola method is used to show that $$\sum_{n\leq x} \sigma(n) =\sum_{n\leq x}\sigma(n)=\frac{\pi^{2}}{12}x^2+O(x\log x),$$ which should be exactly what you are looking for. From here, partial summation yields $$\sum_{n\leq x} n\sigma(n) =\frac{\pi^{2}}{18}x^3+O(x^2\log x).$$

I will post Part II soon which proves that the error term is not $o(x^2\log \log x),$ and oscillates from negative to positive with a magnitude of $x^2\log \log x$ infinitely often.

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I think the error term in Theorem 2 (Part I) is not what you intended (though it is correct...). –  Charles Jun 28 '12 at 2:23

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