# Lower Bound on “exponential” sum

Let $\tau(n)$ be the divisor function. Let $a$ be either a constant, or a function of $X$ that is slowly varying with $X,$ say $X/\log(X)<a(X)<X \log(X),$ for example. I want to lower bound sums of the following form $$\sum_{1\leq n\leq X} a^{1-\frac{\tau(n)}{D}},\quad(1)$$ and $$\sum_{1\leq n\leq X: n\in I} a^{1-\frac{\tau(n)}{D}},\quad(2)$$ where $I$ is an index set of roughly $n/2$ integers. Here, $0<D\leq X$ is also a function of $X$ that is slowly varying with $X.$

We can, of course, factor out an $a$.

I know the divisor function values for the interval $[1,X]$ obeys a kind of arcsine distribution law, and that should help obtain a bound, but I haven't been able to obtain one. Directly applying the AGM (arithmetic-geometric mean inequality) seems to be too weak.

Edit: For (1) we can apply the AGM $$a X\left( X^{-1} \sum_{1\leq n\leq X} a^{-\tau(n)/D}\right)\geq a X\left( \prod_{1\leq n\leq X} a^{-\tau(n)/D}\right)^{1/X}$$ which gives $$a X\left( a^{-\sum_{1\leq n\leq X} \tau(n)/DX}\right)\approx aX \left( a^{-(X \log X +(2\gamma-1)X+o(\sqrt{X}))/DX}\right)= X a^{1-O(\frac{\log X}{D})}.$$ As for 2, the question is how to obtain a worst case bound by using the distribution I alluded to in the original question, plus something else(?).

• Do you mean I has about X/2 integers? – The Masked Avenger Mar 20 '15 at 16:17
• @ The Masked Avenger: Yes. – kodlu Mar 20 '15 at 19:35
• When I pick $I$ to be of cardinality $n/2$ and pick it to give the smallest value for the sum [at each $n$], it seems like the sum $\sum_{1\leq n \leq X: n \in I} \tau(n)$ may well be something like $X \log^2 X$ or $X^{1+\epsilon}$ though I doubt the second one. This was checked on Mathematica for $n\leq 4000$ for now. – kodlu Mar 22 '15 at 9:32
• Oops, should have written $X \log X$ above, read the plot wrong. – kodlu Mar 22 '15 at 9:48

Consider $$\sum_{1\leq n\leq X: n\in I} a^{1-\frac{\tau(n)}{D}},\quad(2)$$ where the question specified $I$ to be an index set of roughly $X/2$ integers. Thus I will take $\mid I\mid \in [(X-t)/2,(X+t)/2]$ for some fixed positive integer $t.$ By using the same argument as for (1) we obtain $$\sum_{1\leq n\leq X: n\in I} a^{1-\frac{\tau(n)}{D}}\geq a \mid I\mid \left( a^{-\sum_{1\leq n\leq X:n\in I} \tau(n)/(D \mid I\mid)}\right)\quad(3).$$ Now focus on the sum in the exponent of (3). Let us assume that we choose $I$ to be an index set (there can be more than one) which achieves the maximal sum of divisors of $n\in I$ subject to the limits on $\mid I\mid$ given above.
If there were a function $f$ growing strictly faster than $O(X \log X)$ such that $$\sum_{1\leq n\leq X: n\in I} \tau(n) \geq f(X),$$ as $X$ grows, then this would imply that $$\sum_{1\leq n\leq X} \tau(n) \geq f(X),$$ since the divisor function is always positive. But this would contradict the fact that the full divisor sum is known to be $O(X \log X).$ Thus the partial sum is at most $O(X \log X).$
Now rewrite (3), using the order estimate from above and the fact that $\mid I\mid =\delta X$ where $\delta>0$ is constant. We get $$\sum_{1\leq n\leq X: n\in I} a^{1-\frac{\tau(n)}{D}}\geq a \delta X \left( a^{-O(X \log X)/(D \delta X)}\right)\gg Xa^{1-O(\frac{\log X}{D})} .$$