- #1
jostpuur
- 2,116
- 19
Fix some constant [itex]0<\alpha \leq 1[/itex], and denote the floor function by [itex]x\mapsto [x][/itex]. The conjecture is that there exists a constant [itex]\beta > 1[/itex] such that
[tex]
\beta^{-n} \sum_{k=0}^{[\alpha\cdot n]} \binom{n}{k} \underset{n\to\infty}{\nrightarrow} 0
[/tex]
Consider this conjecture as a challenge. I don't know how to prove it myself.
It can be proven easily, that if [itex]\beta > 2[/itex], then
[tex]
\beta^{-n} \sum_{k=0}^{[\alpha\cdot n]} \binom{n}{k} \underset{n\to\infty}{\to} 0
[/tex]
so the task is not trivial. My numerical observations suggest that the conjecture is still true, and some beta from the interval [itex]1 < \beta < 2[/itex] can be found so that the claim becomes true.
[tex]
\beta^{-n} \sum_{k=0}^{[\alpha\cdot n]} \binom{n}{k} \underset{n\to\infty}{\nrightarrow} 0
[/tex]
Consider this conjecture as a challenge. I don't know how to prove it myself.
It can be proven easily, that if [itex]\beta > 2[/itex], then
[tex]
\beta^{-n} \sum_{k=0}^{[\alpha\cdot n]} \binom{n}{k} \underset{n\to\infty}{\to} 0
[/tex]
so the task is not trivial. My numerical observations suggest that the conjecture is still true, and some beta from the interval [itex]1 < \beta < 2[/itex] can be found so that the claim becomes true.