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.


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.


2 Answers 2


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) $$\sum_{n\le x}\sigma_v(n)=x^v (\sum_{n\le x}\sigma_v(n) n^{-v})+\sum_{m\le x-1} (\sum_{n\le m}\sigma_v(n) n^{-v})(m^v-(m+1)^v)$$ $$ = \ldots $$

  • $\begingroup$ 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

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:

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ @Gary Thanks for labeling the equations $\endgroup$
    – TravorLZH
    Nov 21, 2020 at 3:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.