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_{5, 1.1} $ represents the 5th term in the sequence where $ c = 1.1 $. Since $ S_{5, 1.1} = 2 $ and $ S_{6, 1.1} = 1 $, this is an example of a decreasing sequence.

For what values of $c$, then, is the sequence non-decreasing?

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!

Edit: My original question asked just for the following two specific cases, which are still the most interesting cases for me, and would still be helpful if a general solution cannot be obtained:

  • 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 \ne 1 $ for which it is non-decreasing?

2 Answers 2


Proof outline:

  1. Define $f(x) = (x+1)^c - x^c$. Obviously $f(n) - 1 \lt S_{n,c} \lt f(n) + 1$. Therefore it is sufficient for $f(x+1) - f(x) \ge 2$ to make sure the sequence is non-descending.
  2. This means that if $f'(x) \ge 2$ the sequence is definitely non-descending. Which means $f'(x) = c(x+1)^{c-1} - cx^{c-1} \ge 2$.
  3. For $c=2$ we actually have an equality, but for $c<2$ the derivative tends to 0, meaning $\lim f'(x) = 0$. Therefore at some point $f'(x) < \epsilon$ and $f(x+1) - f(x) < \epsilon$.
  4. Define $g(n) = S_{n,c}-f(n)$. $f(x)$ tends to infinity, but the derivative tends to $0$. Therefore we may assume that for sufficiently large $n$, $f(x)$ is very nearly constant. In this neighborhood, whenever $g(n) > 0$ but $g(n+1) < 0$ we have a decrease, because $f(n+1) \approx f(n)$. $g(x)$ switches sign infinitely many times, and for all $m$ there exists $n$ such that $g(m+n)$ has a different sign than $g(m)$.
  5. We go to an $m$, where the sign of $g(m)$ is positive but the sign of $g(m+1)$ is negative, and we have our "winner".
  6. So the critical value is $c = 2$. For non-integer $c<2$, $S_{n,c}$ won't be strictly non-decreasing, but for values close enough to $2$ we'll have to go farther and farther towards infinity to find the counter-example.
  7. For $c=1$ we have $S_{n,1} = 1$ and therefore non-decreasing. For $c<1$ we have $S_{n,c} = 0$ for sufficiently large $n$, therefore it can't be non-decreasing.

  8. To summarize:

    • $ c >= 2$ non-decreasing because of the derivative.
    • $ 2 > c > 1 $ is not non-decreasing because of the derivative.
    • $ c = 1$ non-decreasing because of the constant value ($1$).
    • $ 0 < c < 1$ decreasing because $S_{n,c} = 0$ infinitely many times.
    • $ c <= 0$ non-decreasing because of the constant value ($0$).
  • $\begingroup$ I'm having trouble understanding point 4. How is $g(x)$ guaranteed to switch signs infinitely many times, especially given that $c = 1$ is a valid non-decreasing sequence? $\endgroup$
    – DPenner1
    Feb 22, 2016 at 19:22
  • 1
    $\begingroup$ When c = 1 the sign is actually neither positive or negative, because g(x) = 0 always. The thing is, when c is not integer the sign is bound to change because at some point (n+1)^c mod 1 will be smaller than n^c mod 1. And at some point n^c mod 1 will be smaller than (n+1)^c mod 1. $\endgroup$
    – Alon Navon
    Feb 22, 2016 at 19:32
  • 1
    $\begingroup$ (n+1)^c - n^c grows ever so slowly when c < 2 and n tends to infinity. So on the one hand at some point (n+1)^c - n^c grows by less than 0.00001 each step, but on the other hand (n+1)^c - n^c diverges, so we know that all those little steps must accumulate past 1 at some point. So there definitely are sign flips. $\endgroup$
    – Alon Navon
    Feb 22, 2016 at 19:37
  • 1
    $\begingroup$ For $0<c<1$ don't you mean that $S_{n,c}$ is infinitely often zero? $\endgroup$
    – Erick Wong
    Feb 22, 2016 at 23:59
  • 1
    $\begingroup$ I clarified the statement anyway. But yes, a single index suffices. $\endgroup$
    – Alon Navon
    Feb 23, 2016 at 0:02

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 $10^7$

  • $\begingroup$ Thanks for the tighter bounds. It's starting to look like $ x = 2 $, but that's just a conjecture. $\endgroup$
    – DPenner1
    Mar 27, 2013 at 0:35
  • $\begingroup$ @DPenner: For what it is worth, nothing found at $n=1.965$ under $50,000,000$ $\endgroup$ Mar 27, 2013 at 0:46

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