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For integers $a,b,k$, how would one evaluate (or find an upper/lower bound) to $$\sum_{n≤k}\frac{\sigma(n)^a}{n^b}$$ Where $\sigma(n)$ denotes the divisor sigma function?

$\bf{Edit:}$ More specifically, how would one evaluate $$\sum_{n≤k}\frac{(\sigma(n)-n)^s}{n^{s-1}}$$ for some integer $s>2$?

RSpeciel
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1 Answers1

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It is well-known that: $$ \sum_{n\leq x}\sigma(n) = \frac{\pi^2}{12} x^2 + O(x)\tag{1} $$ and the average order of $\sigma(n)^a$ can be derived from Euler's product and tauberian theorems.
For instance

$$\begin{eqnarray*}\sum_{n\geq 1}\frac{\sigma(n)^2}{n^s}&=&\prod_{p}\left(1+\frac{(p+1)^2}{p^s}+\frac{(1+p+p^2)^2}{p^{2s}}+\ldots\right)\\&=&\prod_p\frac{1-\frac{1}{p^{2s-2}}}{\left(1-\frac{1}{p^s}\right)\left(1-\frac{1}{p^{s-1}}\right)^2\left(1-\frac{1}{p^{s-2}}\right)}\\&=&\frac{\zeta(s)\zeta(s-1)^2\zeta(s-2)}{\zeta(2s-2)}\tag{2}\end{eqnarray*}$$ leads to $$ \sum_{n\leq x}\sigma(n)^2 = \frac{5\zeta(3)}{6} x^3+O(x^2)\tag{3} $$ and $\sum_{n\leq x}\frac{\sigma(n)^2}{n^b}$ can be estimated through summation by parts. Of course the behaviour of $\sum_{n\leq x}\frac{\sigma(n)^a}{n^b}$ strongly depends on $b>(a+1)$ (leading to a convergent series) $b=(a+1)$ (leading to a slowly divergent series) or $b<(a+1)$ (leading to a polynomial bound).

Jack D'Aurizio
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  • See edit... I'm trying to evaluate the sum in the edit by expanding the polynomial in the numerator and obtaining asymptotic approximations for each term as $k$ tends to infinity – RSpeciel Jul 07 '17 at 18:10
  • @Romain: if I did not mess up, the above technique leads to $$\sum_{n\leq x}\frac{(\sigma(n)-n)^2}{n}=\frac{6-2\pi^2+15\zeta(3)}{12}x^2+O(x)$$ for $s=2$, but computations are not that pleasant for a generic $s>2$. – Jack D'Aurizio Jul 07 '17 at 18:47
  • thank you very much for the answer. For clarification, how did you use hardy littlewoods's tauberian theorem to get from (2) to (3)? – RSpeciel Jul 07 '17 at 18:50
  • @Romain: I used the tauberian theorem to derive $(3)$ from $(2)$: the abscissa of convergence of the series in $(2)$ is $\sigma=3$ and by the asymptotic behaviour of $\frac{\zeta(s)\zeta(s-1)^2\zeta(s-2)}{\zeta(2s-2)}$ in a right neighbourhood of $s=3$ the statement $(3)$ follows. – Jack D'Aurizio Jul 07 '17 at 18:53