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I am looking for the tightest known bound for the sum
$$\sum_{\substack{1\leq k\leq j^\alpha \\ k\mid j}}k^\lambda$$ where $j$ is a large positive integer, $\alpha\in(0,1)$ and $\lambda\geq 1$.

I am also interested whether it is possible to give any better estimate on the sum
$$\sum_{j=n}^{n^2} \frac{1}{j^\lambda} \sum_{\substack{1\leq k\leq j^\alpha \\ k\mid j}}k^\lambda$$ with the same parameters $\alpha, \lambda$ and for some large positive integer $n$ (if it makes any difference one can think on $n$ as being a large prime number), rather than just using the bound for the previous sum.

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  • $\begingroup$ You might find this paper useful: Norton, Karl K. "Upper bounds for sums of powers of divisor functions." Journal of number theory 40.1 (1992): 60-85.‏ It obtains a quite tight upper bound for the first sum. $\endgroup$
    – assaferan
    Commented Feb 12, 2018 at 14:54
  • $\begingroup$ @assaferan are you sure? I don't see anything of that form in the paper you mentioned. Could you please be more specific as to where in the paper this estimate appears? $\endgroup$
    – Itay
    Commented Feb 12, 2018 at 15:20
  • $\begingroup$ Oh, my apologies. In this paper he only proves bounds for similar expressions with $\zeta(s)^2$. But it seems to me that one can obtain similar (to Theorem 1.7) results by using the same methods with $\zeta(s)\zeta(s-a)$, which then will apply to your case as well. $\endgroup$
    – assaferan
    Commented Feb 14, 2018 at 10:33

2 Answers 2

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The second sum, can be rewritted as $$\sum_{k=1}^{n^{2\alpha}} k^\lambda \sum_{j=\max\{n,k^{1/\alpha}\}\atop k\mid j}^{n^2} \frac{1}{j^\lambda} = \sum_{k=1}^{n^{2\alpha}} k^\lambda \sum_{\ell=\lceil \max\{n/k,k^{1/\alpha-1}\}\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{(k\ell)^\lambda} = \sum_{k=1}^{n^{2\alpha}} \sum_{\ell=\lceil \max\{n/k,k^{1/\alpha-1}\}\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda}$$ $$=\sum_{k=1}^{n^{\alpha}}\sum_{\ell=\lceil n/k\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda} + \sum_{k=n^{\alpha}}^{n^{2\alpha}} \sum_{\ell=\lceil k^{1/\alpha-1}\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda}.$$ The inner sums can be bounded by the integrals: $$\int_{L+1}^{n^2/k} \frac{d\,t}{t^\lambda} \leq \sum_{\ell=\lceil L\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda} \leq \int_{L-1}^{n^2/k} \frac{d\,t}{t^\lambda}.$$ E.g., for $\lambda>1$, we have $$C_1k^{\lambda-1} \leq \sum_{\ell=\lceil n/k\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda} \leq C_2k^{\lambda-1},$$ where $$C_1=\frac{n^{\lambda-1}-1}{(\lambda-1)n^{2(\lambda-1)}}, \qquad C_2= \frac{(n-n^\alpha)^{1-\lambda}-1}{(\lambda-1)n^{2(\lambda-1)}}.$$ Then $$\sum_{k=1}^{n^{\alpha}} \sum_{\ell=\lceil n/k\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda} \geq \sum_{k=1}^{n^{\alpha}} C_1 k^{\lambda-1}\geq \frac{C_1}{\lambda} n^{\alpha\lambda}$$ and $$\sum_{k=1}^{n^{\alpha}} \sum_{\ell=\lceil n/k\rceil}^{\lfloor n^2/k\rfloor} \frac{1}{\ell^\lambda} \leq \sum_{k=1}^{n^{\alpha}} C_2 k^{\lambda-1}\leq \frac{C_2}{\lambda} (n^\alpha+1)^{\lambda}.$$ The other sum is estimated similarly.

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For the purposes of calibration, note that the Rankin-trick inequality $$\sum_{\substack{1\leq k\leq j^\alpha \\ k\mid j}}k^\lambda \le \sum_{k\mid j} k^\lambda \bigg( \frac{j^\alpha}k \bigg)^\lambda$$ gives the upper bound $j^{\alpha\lambda}\tau(j)\ll_\varepsilon j^{\alpha\lambda+\varepsilon}$ for the first sum (where $\tau(j)$ is the number of divisors of $j$); and this is reasonably tight, since the sum can be at least $(j^\alpha)^\lambda$ when $j^\alpha$ is in fact a divisor of $j$.

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