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T. Amdeberhan
T. Amdeberhan
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You're right. It is easy with Stirling's approximation $$N! \sim \sqrt{2\pi N}\left(\frac{N}e\right)^N$$ so that $$\frac{(2N)!}{N!N!2^{2N}} \sim \frac{\sqrt{2\pi(2N)}(2N)^{2N}}{e^{2N}2^{2N}}\frac{e^{2N}}{(2\pi N)N^{2N}}=\frac1{\sqrt{\pi N}}\rightarrow 0$$$$L_N=\frac{(2N)!}{N!N!2^{2N}} \sim \frac{\sqrt{2\pi(2N)}(2N)^{2N}}{e^{2N}2^{2N}}\frac{e^{2N}}{(2\pi N)N^{2N}}=\frac1{\sqrt{\pi N}}\rightarrow 0$$ as $N\rightarrow\infty$.

You're right. It is easy with Stirling's approximation $$N! \sim \sqrt{2\pi N}\left(\frac{N}e\right)^N$$ so that $$\frac{(2N)!}{N!N!2^{2N}} \sim \frac{\sqrt{2\pi(2N)}(2N)^{2N}}{e^{2N}2^{2N}}\frac{e^{2N}}{(2\pi N)N^{2N}}=\frac1{\sqrt{\pi N}}\rightarrow 0$$ as $N\rightarrow\infty$.

You're right. It is easy with Stirling's approximation $$N! \sim \sqrt{2\pi N}\left(\frac{N}e\right)^N$$ so that $$L_N=\frac{(2N)!}{N!N!2^{2N}} \sim \frac{\sqrt{2\pi(2N)}(2N)^{2N}}{e^{2N}2^{2N}}\frac{e^{2N}}{(2\pi N)N^{2N}}=\frac1{\sqrt{\pi N}}\rightarrow 0$$ as $N\rightarrow\infty$.

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T. Amdeberhan
T. Amdeberhan
  • 43.2k
  • 5
  • 57
  • 217

You're right. It is easy with Stirling's approximation $$N! \sim \sqrt{2\pi N}\left(\frac{N}e\right)^N$$ so that $$\frac{(2N)!}{N!N!2^{2N}} \sim \frac{\sqrt{2\pi(2N)}(2N)^{2N}}{e^{2N}2^{2N}}\frac{e^{2N}}{(2\pi N)N^{2N}}=\frac1{\pi\sqrt{2}N}\rightarrow 0$$$$\frac{(2N)!}{N!N!2^{2N}} \sim \frac{\sqrt{2\pi(2N)}(2N)^{2N}}{e^{2N}2^{2N}}\frac{e^{2N}}{(2\pi N)N^{2N}}=\frac1{\sqrt{\pi N}}\rightarrow 0$$ as $N\rightarrow\infty$.

You're right. It is easy with Stirling's approximation $$N! \sim \sqrt{2\pi N}\left(\frac{N}e\right)^N$$ so that $$\frac{(2N)!}{N!N!2^{2N}} \sim \frac{\sqrt{2\pi(2N)}(2N)^{2N}}{e^{2N}2^{2N}}\frac{e^{2N}}{(2\pi N)N^{2N}}=\frac1{\pi\sqrt{2}N}\rightarrow 0$$ as $N\rightarrow\infty$.

You're right. It is easy with Stirling's approximation $$N! \sim \sqrt{2\pi N}\left(\frac{N}e\right)^N$$ so that $$\frac{(2N)!}{N!N!2^{2N}} \sim \frac{\sqrt{2\pi(2N)}(2N)^{2N}}{e^{2N}2^{2N}}\frac{e^{2N}}{(2\pi N)N^{2N}}=\frac1{\sqrt{\pi N}}\rightarrow 0$$ as $N\rightarrow\infty$.

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T. Amdeberhan
T. Amdeberhan
  • 43.2k
  • 5
  • 57
  • 217

You're right. It is easy with Stirling's approximation $$N! \sim \sqrt{2\pi N}\left(\frac{N}e\right)^N$$ so that $$\frac{(2N)!}{N!N!2^{2N}} \sim \frac{\sqrt{2\pi(2N)}(2N)^{2N}}{e^{2N}2^{2N}}\frac{e^{2N}}{(2\pi N)N^{2N}}=\frac1{\pi\sqrt{2}N}\rightarrow 0$$ as $N\rightarrow\infty$.