Timeline for Upper limit on the central binomial coefficient
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jan 2, 2021 at 16:49 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
added 281 characters in body
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| Jan 1, 2021 at 15:08 | comment | added | Fedor Petrov | Warning: $1/\sqrt{\pi(n+1/4)}$ is an upper bound | |
| Jan 1, 2021 at 14:41 | comment | added | aorq | Yes, I know that it's tight in the weakest possible sense, namely equality for $n=0$ and poor otherwise. Shortly after I wrote my comment, I saw that $1/\sqrt{\pi(n+1/4)}$ was tighter in a more meaningful sense (on the other side), namely ever better as $n\to\infty$. Thanks for your explanation! | |
| Jan 1, 2021 at 9:51 | comment | added | Fedor Petrov | @aorq It is not so much tight. The sequence $4^{-n}{2n\choose n}\sqrt{n+c}$ eventually decreases if $c>1/4$ (this is straightforward: divide the squares of two consecutive terms and subtract 1) and its limit equals $1/\sqrt{\pi}$. Thus for large enough $n$ we get the upper bound $1/\sqrt{\pi (n+c)}$. For $c=1/\pi$ this just works from the very beginning. | |
| Jan 1, 2021 at 2:54 | comment | added | aorq | Nice proof! Is there an easy way to get the tight lower bound $1/\sqrt{\pi n+1}$? It has such a nice form.. | |
| Jan 1, 2021 at 1:49 | history | answered | Fedor Petrov | CC BY-SA 4.0 |