What is the tighest upper and lower bounds for $\log{ (\binom{n}{n/2})}$? I know that a naive upper bound would be $n \log{n}$ but is there any tighter bound? If so, how do I approach to solve this?
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Stirling? ${}{}$ – Maximilian Janisch Jul 22 '20 at 15:13
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Does this help? (Only related related occurred in approach 0 on searching by one format) – UmbQbify Jul 22 '20 at 15:15
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@MaximilianJanisch probably I should do that. – am_rf24 Jul 22 '20 at 15:35
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@UmbQbify - That still leaves it at the same point. But as the post suggests, maybe Stirling would do. – am_rf24 Jul 22 '20 at 15:35