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Question:

Define the sum of $v$-powers of divisor $\sigma_v(n)=\sum_{d|n}d^v$ for $v \in \mathbb{R}$.
Prove that for all $v>0$, $$\frac{ \sum_{i=1}^{n}\sigma_v(i)}{n}\sim\frac{n^v\zeta(v+1)}{v+1}$$ where $\zeta$ is the Riemann zeta function.

Attempts:

By using the convolution $\sigma_1=\mathbb{1}*\text{Id}$, (where $\mathbb{1}(n)=1$, $\text{Id}(n)=n$)
we have $\displaystyle \sum_{i=1}^n\sigma_v(i)=\sum_{d=1}^{n}\sum_{k\leq \frac{n}{d}}k^v$.
When $v=1$, we know the sum $\displaystyle \sum_{k\leq \frac{n}{d}}k = \sum_{k=1}^{[\frac{n}{d}]}k=\frac{1}{2}([\frac{n}{d}]+1)[\frac{n}{d}]$,
then by bounding $\frac{n}{d}-1<[\frac{n}{d}]\leq\frac{n}{d}$,
we can show $\displaystyle \sum_{i=1}^n\sigma_1(i)=\frac{n^2}{2}\sum_{d=1}^n\frac{1}{d^2}+f(n)$ for some $|f(n)|<n\log{n}$ , when $n\geq 4$.
The "2" in $d^2$, $n^2$, $\frac{1}{2}$ comes from the rather easy formula of sum of positive integers.
However, when $v\neq 1$, the term becomes the sum of power of integers and I cannot easily find a formula for it.
The closest I have heard of is Faulhaber's_formula with Bernoulli numbers as coefficients, but the coefficients are hard to handle and the formula seems to be for integer power only.

Another way I think of is by inequality. Power-Mean Inequality looks usable, but it does not have the ratio $\displaystyle\frac{1}{v+1}$ so it does not seems to be the way.

The third way is possibly constructing the general case from the result of $\sigma_1$, but I don't know how to either.

Any help is very much appreciated.

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2 Answers 2

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I don't think you'll do it this way, Euler-McLaurin (the extension of Faulhaber to non-integer $v$) gives a good approximation of $\sum_{k\le y} k^v$ but as a function of $\lfloor y\rfloor$ not $y$.

$$\sum_{n\le x}\sigma_v(n) n^{-v}= \sum_{n\le x}\sigma_{-v}(n) =\sum_{k\le x}k^{-v} \lfloor x/k\rfloor = \sum_{k\le x}k^{-v} (x/k+O(1))$$ $$= x (\zeta(v+1)+O(x^{-v}))+ O(1)+O(x^{1-v})$$

Then by partial summation (for $x$ integer) $$\sum_{n\le x}\sigma_v(n)=x^v (\sum_{n\le x}\sigma_v(n) n^{-v})+\sum_{m\le x-1} (\sum_{n\le m}\sigma_v(n) n^{-v})(m^v-(m+1)^v)$$ $$ = \ldots $$

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  • $\begingroup$ Thank you for your answer! After working on it for some time I still find it difficult to understand some parts. Do you mind explaining (1) How to go from $\sum_{n\le x}\sigma_{-v}(n)$ to $\sum_{k\le x}k^{-v} \lfloor x/k\rfloor$? In particular, why does the summation index change? (2) How to get the 2 Big-O's from $\sum_{k\le x}k^{-v} (x/k+O(1))$ to $ x (\zeta(v+1)+O(x^{-v}))+ O(x^{1-v})$ ? (3) In $= \ldots$, there are a lot of Big-O's, how should we proceed? $\endgroup$
    – Wegelip
    Nov 18, 2020 at 1:36
  • $\begingroup$ $\sum_{n\le x}\sigma_{-v}(n)=\sum_{n\le x}\sum_{dk=n}k^{-v}=\sum_{dk\le x} k^{-v}$. And $\sum_{k\le x}k^{-v} (x/k+O(1))=x\sum_{k\le x}k^{-v-1}+O(\sum_{k\le x}k^{-v})$. It gives $x\zeta(v+1)+O(1)+O(x^{1-v})$ which fits nicely into the partial summation. $\endgroup$
    – reuns
    Nov 18, 2020 at 1:37
  • $\begingroup$ Thank you and sorry for asking again. I think I am not fluent in Big-O's enough to understand. Do you mind detailing the steps to break apart $O(\sum_{k\le x}k^{-v})$? (also, is it $x (\zeta(v+1)+O(x^{-v}))+ O(1)+O(x^{1-v})$ or $x \zeta(v+1)+ O(1)+O(x^{1-v})$?) For fitting nicely into the partial summation, I guess it means $m^v$ somehow cancelling with $O(x^{-v})$. I am still not sure how to complete it, could I ask for the explicit steps in $= \dots$? Thank you very much. $\endgroup$
    – Wegelip
    Nov 18, 2020 at 2:23
  • $\begingroup$ If you are worried by those things then simply use that $x\sum_{k\le x}k^{-v-1} \sim x \zeta(v+1)$ and $\sum_{k\le x} k^{-v} = O(\int_1^x y^{-v}dy)=O(x^{\max(0,1-v)})= o(x)$ so that $$\sum_{n\le x}\sigma_v(n) n^{-v} \sim x\zeta(v+1)$$. Then $m^v-(m+1)^v\sim -v m^{v-1}$ and your problem reduces to show that $$\sum_{m\le x-1} (\sum_{n\le m}\sigma_v(n) n^{-v})(m^v-(m+1)^v)\sim \sum_{m\le x-1} m\zeta(v+1) (-v) m^{v-1}\sim \zeta(v+1) \frac{-v}{v+1}x^{v+1}$$ $\endgroup$
    – reuns
    Nov 18, 2020 at 2:28
  • $\begingroup$ @reuns It does not have to be that complicated. Analytic number theory had already provided us with some useful asymptotic tools on summing powers $\endgroup$
    – TravorLZH
    Nov 20, 2020 at 12:12
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Using techniques like Riemann-Stieltjes integration, we can show that

$$ \sum_{k\le n}{1\over k^s}=\zeta(s)+\mathcal O\left(1\over n^{s-1}\right)\quad \Re(s)>1\tag1 $$ $$ \sum_{k\le n}\frac1k=\log n+\gamma+\mathcal O\left(\frac1n\right)\tag2 $$ $$ \sum_{k\le n}k^a={k^{a+1}\over a+1}+\mathcal O(k^a)\quad a\ge0\tag3 $$

and further works on this problem solely depend on these identities.

First, let's exchange the order of summation

$$ \begin{aligned} \frac1n\sum_{k\le n}\sigma_v(k) &=\frac1n\sum_{k\le n}\sum_{d|k}d^v =\frac1n\sum_{qd\le n}d^v \\ &=\frac1n\sum_{q\le n}\color{green}{\sum_{d\le n/q}d^v} \end{aligned} $$

To continue, we apply (3) to the green part:

$$ \begin{aligned} \frac1n\sum_{k\le n}\sigma_v(k) &=\frac1n\sum_{q\le n}\left[{n^{v+1}\over(v+1)q^{v+1}}+\mathcal O\left(n^v\over q^v\right)\right] \\ &={n^v\over v+1}\color{orange}{\sum_{q\le n}{1\over n^{v+1}}}+\mathcal O\left(n^{v-1}\color{blue}{\sum_{q\le n}{1\over q^v}}\right) \end{aligned} $$

Since $v>0$, we can apply (1) to the orange sum. Because it is not known whether $v>1$, it is better for us to consider each different situations for the blue sum:

  • For $v>1$, we know that the blue series converges, so it becomes $\mathcal O(n^{v-1})$.
  • For $v=1$, we can apply (2) to get $\mathcal O(n^{v-1}\log n)$.
  • For $0<v<1$, we can use (3) to obtain $\mathcal O(1)$.

In each of the aforesaid situations, we observe that the error term would not exceed $n^v$, leaving us this asymptotic relation:

$$ \frac1n\sum_{k\le n}\sigma_v(k)\sim{n^v\zeta(v+1)\over v+1} $$

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  • $\begingroup$ Yes tks, for $v > 0$ and $r\in (0,\min(1,v))$, $$\sum_{q\le n} \sum_{d\le n/d} d^v= \sum_{q\le n} (n/q)^{v+1}/(v+1)+O((n/q)^v)= n^{v+1} (\zeta(v+1)/(v+1)+o(1))+O(n^{v+1-r})$$ works fine while $\sum_{d\le n} d^v \lfloor n/d\rfloor$ doesn't $\endgroup$
    – reuns
    Nov 20, 2020 at 15:41
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    $\begingroup$ @Gary Thanks for labeling the equations $\endgroup$
    – TravorLZH
    Nov 21, 2020 at 3:56

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