How to bound $\sum_{1 \leq x_1, ..., x_n \leq N} lcm(x_1, ..., x_n)^{- \delta}$?

Question

Let $\delta>0$. I am interested in obtaining a bound for the sum $\sum_{1 \leq x_1, ..., x_n \leq N} \operatorname{lcm}(x_1, ..., x_n)^{- \delta}$ where lcm denotes the lowest common multiple of the numbers. I would appreciate any comments and suggestions! Thank you very much.

Answer 1

For $\delta>1$ the sum is bounded, while for $\delta=1$ it grows by a power of $\log N$.

So let me focus on $0<\delta<1$ and $N\geq 2$. For a simple lower bound, we have $$ \sum_{1 \leq x_1, ..., x_n \leq N} \operatorname{lcm}(x_1, ..., x_n)^{- \delta}\geq \left(\sum_{1 \leq x \leq N} x^{- \delta}\right)^n\gg_{n,\delta}N^{n(1-\delta)}. $$ For a simple upper bound, we have $$ \sum_{1 \leq x_1, ..., x_n \leq N} \operatorname{lcm}(x_1, ..., x_n)^{- \delta}\leq \sum_{1\leq m\leq N^n}\frac{\tau(m)^n}{m^\delta}\ll_{n,\delta}N^{n(1-\delta)}(\log N)^{2^n-1}.$$ So, up to a logarithmic factor, the sum grows like $N^{n(1-\delta)}$.

Answer 2

Here is a start on the problem. For each value k of lcm in your sum, you find how many tuples M_k give that lcm, and then you sum M_k(k^{-delta}) over enough values of k, which includes values up to N and some smattering up to near N^n.

Let's say k is a prime power, p^e , so k is less than N. Then the number of tuples is seen to be (e+1)^n -e^n. So already we have about N/log N values of k solved.

Now let's say k is the proper product of two prime powers. We get an upper bound of ((e+1)^n - e^n)((f+1)^n - f^n), where I bet you can guess what f is. For a more precise enumeration, you have to toss out numbers of the form p^cq^d that are larger than N, or decide that they are small enough to include so that you have an upper bound.

This extends to k being a product of more than one prime factor, but one has to watch for counting things outside the sum. However, as k gets large, k^{-delta} may get small enough that it won't hurt to include additional terms. The major contribution will come from those k at most N anyway, and provide an informative lower bound.

Gerhard "That's It, More Or Less" Paseman, 2017.12.19.