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Can the following claim be extended for some complex $k$. Perhaps: all $k$ with real part greater than or equal to $1$? Do the arguments below fall apart for complex $k$ for some reason?

Claim. For any real $k\ge1$ $$\lim_{x\to\infty}\frac{1}{x^{k+1}}\sum_{n=1}^x \sigma_k(n)=\frac{\zeta(k+1)}{k+1}$$

Where $\sigma_k(n)=\sum_{d|n}d^k$ and $\zeta(s)=\sum_{n=1}^\infty{\frac{1}{n^s}}$. The case for $k=1$ is shown here and the two lemmata below will extend the claim for $k>1$. I really just copied the arguments from the linked post and added in this symbol $k$.

Lemma 1. For $k>1$ we have $$\sum_{n=1}^x \frac{\sigma_k(n)}{n^k}\in\zeta(k+1)x+O(1)$$

Aside I should comment that I have some feelings about big O notation. I think we should not obfuscate the 'is' of identity and the 'is' of predication. This has caused problems in the past... We should really say $f\in O(x^3)$ and not $f=O(x^3)$. This opinion is not unique to me. And I don't want to start a discussion of notation here. But I am explaining why we see $\in$ and not the more commonly seen $=$ in my presentation of the lemma. It's because of my notation feelings.

Proof $$\begin{align} &\sum_{n=1}^x\frac{\sigma_k(n)}{n^k} \\ &=\sum_{n=1}^x \frac{1}{n^k}\sum_{d|n} (\frac{n}{d})^k \\ &=\sum_{n=1}^x \sum_{d|n} (\frac{1}{d})^k \\ &=\sum_{n=1}^x \sum_{A\le \frac{x}{d}} (\frac{1}{d})^k \\ &=\sum_{d=1}^x (\frac{1}{d})^k \sum_{A\le \frac{x}{d}}1 \\ &=\sum_{d=1}^x (\frac{1}{d})^k \bigg \lfloor \frac{x}{d} \bigg \rfloor \\ & \text{And since} \bigg \lfloor \frac{x}{d} \bigg \rfloor \in \frac{x}{d}+O(1) \text{we can substitute into the expression above to arrive at} \\ &\sum_{n=1}^x \frac{\sigma_k(n)}{n^k} \in \sum_{n=1}^x{\frac{1}{d^k}}{\bigg(\frac{x}{d}+O(1) \bigg)} \\ &\subseteq x\sum_{d=1}^x\frac{1}{d^{k+1}}+O\bigg( \sum_{d=1}^x\frac{1}{d^k} \bigg) \\ &\subseteq x\bigg(\zeta(k+1)+O\big(\frac{1}{x}\big) \bigg)+O(x^{1-k})\\ &\subseteq x\zeta(k+1)+O(1) \end{align} $$

This is justified because for $k \geq 1$ we have $O(x^{1-k}) \subseteq O(1)$. This concludes the first lemma. $\square$

Lemma 2. For $k>1$, $$\sum_{n=1}^x\sigma_{k}(n) \in \frac{\zeta(k+1)}{k+1}x^{k+1}+O(x^k)$$

Proof We will use Abel's Summation:

$$\sum_{n=1}^x a_nf(n)= A(x)f(x) +\int_1^x A(t)f'(t) dt $$ where $A(x)=\sum_{n=1}^x a_n $. We will take $a_n=\frac{\sigma_k(n)}{n^k}$ and $f(x)=x^k \implies f'(x)=kx^{k-1}$.

Substituting we have

$$\sum_{n=1}^x \sigma_k(n)= \sum_{n=1}^x \bigg(\frac{\sigma_k(n)}{n^k} \bigg){n^k}=x^k \sum_{n=1}^x \frac{\sigma_k(n)}{n^k}-k\int_1^x{t^{k-1}\sum_{n=1}^t{\frac{\sigma_k(n)}{n^k}}dt}$$

But in view of Lemma 1 we can see that this means

$$\begin{align} &\sum_{n=1}^x \sigma_k(n) \\ &\in x^k[\zeta(k+1)x+O(1)]-k\int_1^xt^{k-1}[\zeta(k+1)t+O(1)]dt\\ &\subseteq\zeta(k+1)x^{k+1}+O(1)-k\zeta(k+1) \int_1^x{t^k}dt+O\bigg(\int_1^x{t^{k-1}dt}\bigg) \\ &\subseteq \zeta(k+1)x^{k+1}+O(1)-k\zeta(k+1)\bigg[\frac{x^{k+1}-1}{k+1}\bigg]+O\bigg(\frac{x^k-1}{k}\bigg) \\ & \subseteq \zeta(k+1) \bigg[ \frac{(k+1)x^{k+1}-kx^{k+1}}{k+1} \bigg]+O(x^k) \\ & \subseteq \frac{\zeta(k+1}{k+1}x^{k+1}+O(x^k) \end{align} $$ This concludes the second lemma. $\square$

The claim above follows from these lemmata.

Mason
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  • Where are you stuck in extending to $\Re(k) > 0$ ? For $\Re(k) < 0$ it is not very different, or you can use that $\sigma_k(n) = n^k \sigma_{-k}(n)$. – reuns Nov 05 '18 at 20:00
  • @reuns. Do you mean $0$ in the comment above? My question what about $R(k)\geq 1$. I really only was thinking about this for boring old zeta before we extend through analytic continuation (which is a concept I don't really grasp yet). Just $\sum_{n=1}^\infty\frac{1}{n^s}$ which is well defined for all $s$ with real part greater than or equal to $1$. My question is just about whether the techniques above work for complex $k$ because I originally contrived the argument thinking about real $k$. I think you're hinting that we can exploit this even further but that isn't really my question. – Mason Nov 05 '18 at 20:48
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    Where are you stuck in extending to $\Re(k) > 1$ ? – reuns Nov 05 '18 at 20:51
  • Where could this go wrong for $k$ in the complex numbers? Hmmm. Lemma 1 seems fine. For lemma 2: Do the calculus techniques work the way I would expect them to for complex $k$? We still have $\int x^k =x^{k+1}/(k+1)$. Right? If so then as far as I can tell there are no issues extending this claim to the complex numbers. – Mason Nov 05 '18 at 20:58
  • @reuns. Are you suggesting that the claim is simple true for all complex numbers? – Mason Nov 05 '18 at 21:01
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    Your lemma $1$ is true for $\Re(k) > 0$ and replacing $O(x^k)$ by $O(x^{\max(k,0)})$ your lemma $2$ is true for $\Re(k) > 0$. For $\Re(k) < 0$ use $\sigma_k(n) = n^k \sigma_{-k}(n)$. For $\Re(k) =0$, $\zeta(s)\zeta(s+k)$ has two poles on $\Re(s)=1$ so lemma 2 has an additional term and lemma 1 is false. – reuns Nov 05 '18 at 21:04
  • Thank you! @reuns. – Mason Nov 05 '18 at 21:11
  • These appear in wiki – Mason Dec 08 '18 at 17:44
  • And I think we can continue to generalize. For a dirichlet character $\chi$ we have $\sum_{n=1}^x \sum_{d|n} \chi(n/d)d^s \in\frac{L(\chi,s+1)}{s+1}x^{s+1} + O(x^s)$ – Mason Mar 10 '21 at 14:20
  • Related argument: https://math.stackexchange.com/questions/3656581/showing-that-sum-m-1n-sigmam-sum-k-1nk-cdot-left-lfloor-frac?rq=1 – Mason Apr 22 '21 at 14:33

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