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Please help me to solve this problem. I can find almost no clue regarding the log part. I tried to break the $\binom{2n}{n}$ part, but in vague...will the breaking help me in anyway?

achille hui
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  • Think about a Riemann sum – marwalix Apr 16 '15 at 20:49
  • Are you sure it's that? Something tells me it's the limit of $\frac{\binom {2n}{n}}{2n}.$ – zhw. Apr 16 '15 at 20:51
  • $\binom{2n}{n}$ doesn't matter. As long as it is an integer $\ge 2$, $\frac{1}{2n\log\binom{2n}{n}} \le \frac{1}{2\log(2) n}$... – achille hui Apr 16 '15 at 20:52
  • Hmm... looking at the edit revisions. the edit from version 1 to version 2 seems wrong. The very original question can be interpreted as finding the limit of $\frac{1}{2n}\log\binom{2n}{n}$ too. – achille hui Apr 16 '15 at 20:56
  • I meant $\frac{\log \binom {2n}{n}}{2n}.$ – zhw. Apr 16 '15 at 20:57
  • all - I changed the question to what it supposed to ask (based on edit version 1) – achille hui Apr 16 '15 at 21:00
  • I think like @zhw it should be $\frac{\log{\binom{2n}{n}}}{2n}$ otherwise the limit is $0$ like in the hints I the answer of Elaqqad – marwalix Apr 16 '15 at 21:06
  • It would be nice if MATHS would join in and reveal which question is being asked rather than those speculations! – String Apr 16 '15 at 21:14

3 Answers3

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$$\log\binom{2n}{n}=\log\prod_{k=1}^{n}\frac{n+k}{k}=\sum_{k=1}^{n}\log\frac{1+\frac{k}{n}}{\frac{k}{n}}\approx n\int_{0}^{1}\log\left(\frac{x+1}{x}\right)\,dx = 2n\log 2.$$

Jack D'Aurizio
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There are $2^{2n}=4^n$ subsets of a set of size $2n$, each of which has between $0$ and $2n$ elements. Of these, ${2n}\choose{n}$ have exactly $n$ elements, and ${{2n}\choose{n}} > {{2n}\choose{k}}$ for any $k\neq n$. Therefore, $$ \frac{4^n}{2n+1} < {{2n}\choose{n}} \le 4^n, $$ so $$ n\log 4-\log(2n+1) < \log{{2n}\choose{n}} \le n \log 4, $$ and $$ \log 2-\frac{\log(2n+1)}{2n} < \frac{1}{2n}\log{{2n}\choose{n}} \le \log 2. $$ We conclude that $$ \lim_{n\rightarrow\infty}\frac{1}{2n}\log{{2n}\choose{n}} = \log 2. $$

mjqxxxx
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Hint You can use the following inequality: $$\frac{4^n}{n}\leq{2n \choose n}\leq 4^n$$ for large $n\geq 4$ (this is the proof for the left inequality)

Elaqqad
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