Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

I recently solved a practical sequence problem, but got curious and tried to generalize it. Let

$$ S_{n, c} = \lfloor \left( n+1 \right)^c \rfloor - \lfloor n^c \rfloor $$

be a set of sequences where $ n \in \mathbb{N}, c \in \mathbb{R}^+ $, that is the positive reals. I'm not sure if my notation is clear, but as an example, $ S_{4, 1.5} $ represents the 4th term in the sequence where $ c = 1.5 $. I have two related questions on this set of sequences:

  • What is the smallest $ x \in \mathbb{R}^+ $ such that for all sequences $ S_{n, c} $ where $c \ge x, S_{n, c} $ is non-decreasing (if it exists)?
  • What is the smallest $ c $ for which it is non-decreasing? (Edit: I'm looking an interesting result excluding the possibility that $ c = 1 $)

An example of a sequence that has decreasing terms is one with $ c = 1.1 $. $ S_{5,1.1} = 2 $, while $ S_{6,1.1} = 1 $.

I've only ever had one class on sequences and series, and its been two years, so I'm quite rusty! Using the binomial theorem, I was able to prove that it's non-decreasing when $ c $ is an integer, but I've so far not been able to extend that. Any help is greatly appreciated!

share|cite|improve this question

For the first bullet, $1.96 \lt x \le 2$, as $S_{7958491,1.96}=8261400,S_{7958492,1.96}=8261399$ and for $c \gt 2$ the difference without the floor signs is $\gt 2$. My program gets slow checking more than $10^7$

share|cite|improve this answer
Thanks for the tighter bounds. It's starting to look like $ x = 2 $, but that's just a conjecture. – DPenner1 Mar 27 '13 at 0:35
@DPenner: For what it is worth, nothing found at $n=1.965$ under $50,000,000$ – Ross Millikan Mar 27 '13 at 0:46

Your Answer


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.