MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
    
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
2  
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 $y\leq\frac{x}{2x-1}.$

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.