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For $x \gt 0,$ what is the greatest $y$ such that $$\sum_ {1\le h^x \le k^y} \frac{1}{h^x k^y}= \infty ?$$

I don't know of any references or methods for this -- not even for $x=1$, for which the series is $$\sum_ {k=1}^{\infty} \frac{H(k)}{k^y},$$ where $H(k)$ is a harmonic number.

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Do I interpret your notation correctly, by assuming that you sum over $k$ from 1 to infinity, and within that summation you sum over $h$ from 1 to an $h_{*}$ defined by $h_{*}^{x}=\left\lfloor k^{y} \right\rfloor$ ? – Johannes Trost Jan 19 at 15:02
$H(k) = \log k + O(1)$ so for $x=1$, the series will diverge for $y \le 1$, and converge for $y \gt 1$. It would not affect the answer, but shouldn't that be $H(k^y)$? – Douglas Zare Jan 19 at 15:04
Interesting to whom? – Asaf Karagila Jan 19 at 21:40
up vote 10 down vote accepted

In short, $$ \begin{cases} \text{when }1\leq x & \text{series diverges when }y\le1\\ \text{when }\frac{1}{2}<x<1 & \text{series diverges when }y\leq\frac{x}{2x-1}\\ \text{when }0<x\leq\frac{1}{2} & \text{series always diverges.} \end{cases} $$

When $x>1$, the inner sum of $$\sum_{k=1}^{\infty}\frac{1}{k^{y}}\sum_{h^{x}\leq k^{y}}\frac{1}{h^{x}}$$ converges so this series diverges precisely when $y\leq 1$. When $x=1$, since $H(k^{y})\sim y\log k$, again we see that this diverges for $y\leq 1$. Lastly, when $0<x<1$ we have that $$\sum_{h^{x}\leq T}\frac{1}{h^{x}}=\int_{1}^{T^{1/x}}\frac{1}{s^{x}}d\left[s\right]\sim\int_{1}^{T^{1/x}}\frac{1}{s^{x}}ds\sim\frac{1}{1-x}\left(T^{1/x}\right)^{1-x}=\frac{1}{1-x}T^{1/x-1},$$ and so the convergence of the series depends on the convergence of $$\sum_{k=1}^{\infty}\frac{1}{k^{y}}k^{y\left(\frac{1}{x}-1\right)}=\sum_{k=1}^{\infty}k^{y/x-2y},$$ and this depends on when $\left(\frac{1}{x}-2\right)y<-1.$ Thus it diverges for every value of $y$ when $0<x\leq\frac{1}{2}$ , and for $\frac{1}{2}<x<1$ , it diverges when $y\leq\frac{x}{2x-1}.$

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