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$$\lim_{n \to \infty} \frac{\big((n+1)(n+1)(n+3)......3n\big)^{1/n}}{(n^{2n})^{1/n}}= ?$$

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closed as off-topic by user21820, user91500, Watson, Laurent Duval, user99914 Jun 12 '16 at 10:18

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  • $\begingroup$ So, essentially, $\lim_{n \to \infty} \left[ \frac{(3n)!}{n!(n^{2n})} \right]^{1/n}$? Have you considered using Stirling? (I haven't looked at this closely, but I would guess it would answer this...) $\endgroup$ – Brian Tung Jun 12 '16 at 5:54
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    $\begingroup$ Show you working, why are you stuck? $\endgroup$ – Abhijit A J Jun 12 '16 at 5:55
  • $\begingroup$ i need the answer in the form e^x. What's Stirling? @BrianTung $\endgroup$ – Tejesh Atr Jun 12 '16 at 5:57
  • $\begingroup$ Here's the Wikipedia article on it. $\endgroup$ – Brian Tung Jun 12 '16 at 5:58
  • $\begingroup$ thanks. But what's the answer to the above question? $\endgroup$ – Tejesh Atr Jun 12 '16 at 6:02
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You want to compute $\lim_{n\rightarrow \infty} ((n+1)(n+2)\cdots(3n))^{1/n} /n^2 = \lim_{n\rightarrow \infty} ((3n)!/n!)^{1/n}/n^2$

By Stirling's approximation, this becomes $\lim_{n \rightarrow \infty} (\sqrt{6 \pi n} (3n/e)^{3n} / (\sqrt{2 \pi n} (n/e)^n))^{1/n} / n^2$ which can be simplified to $\lim_{n \rightarrow \infty} (27^n\sqrt{3} (n/e)^{2n})^{1/n} / n^2 = \lim_{n \rightarrow \infty} 27\cdot3^{1/2n}\cdot (n/e)^{2} / n^2 = 27/e^2$.

And here's some numerical verification https://ideone.com/hwHyVI

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  • $\begingroup$ I am unaware of Stirling's approximation. Is there any simpler method? Thanks. $\endgroup$ – Tejesh Atr Jun 12 '16 at 6:16
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(No Stirling's approx.)

Result: If $\lim_{n \to \infty }\frac{a_{n+1}}{a_n}=l$ then $\lim_{n\to\infty} (a_n)^{1/n}=l.$

So all we need to find is just limit of $\frac{a_{n+1}}{a_n}.$

Now given $a_n=\frac{(3n)!}{n!n^{2n}}$ so $a_{n+1}=\frac{(3(n+1))!}{(n +1)!(n+1)^{2(n+1)}}$. So $$\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=\lim_{n\to \infty}\frac{(3(n+1))! n!n^{2n}}{(n+1)!(n+1)^{2(n+1)}(3n)!}\\=\lim_{n\to \infty}\frac{3(n+1)(3n+2)(3n+1)n^{2n}}{(n+1)^3(n+1)^{2n}}\\=\lim_{n\to \infty}\frac{3(3n+2)(3n+1)}{(n+1)^2}\left(\frac{n}{n+1}\right)^{2n}\\=\lim_{n\to \infty}\frac{3(3+\frac{2}{n})(3+\frac{1}{n})}{(1+\frac{1}{n})^2}\left(\frac{n}{n+1}\right)^{2n}=\frac{27}{e^2}$$

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May be you don't need any approximation for this question. You can do it directly.

First, take ln both the sides and arrange it in this way $$lnL=\lim_{n\to\infty}(\sum_{i=1}^{2n}\frac{\ln(i+n)}{n}-\sum_{i=1}^{2n}\frac{\ln(n)}{n})$$ I have arranged the second term in form of a summation. Now combine both of the summations $$\sum_{i=1}^{2n}\frac{\ln(\frac{i}{n}+1)}{n}$$. Now assume $\frac{i}{n}$ as $t$ and $\frac{1}{n}$ as $dt$ . So now this limits is converted to a integration $$lnL=\int_{t=0}^2 ln(1+t)dt$$ Limits are decided by $t=\lim_{n\to\infty}(\frac{i}{n})$ . Put the values of $i=1$ to $i=2n$, you will get the limits. So now integration is $lnL=\int_{t=1}^3 ln(t)dt=3ln3-2$ which means $$L=\frac{27}{e^2}$$ Final answer without any approximation.

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In the same spirit as other answers and comments, let us consider $$A_n= \left[ \frac{(3n)!}{n!(n^{2n})} \right]^{1/n}$$ Taking logarithms $$\log(A_n)=\frac 1 n\left(\log(3n!)-\log(n!)-2n\log(n) \right)$$ Now, using Stirling approximation $$\log(k!)\sim k\log\left(\frac k e\right)+\frac 12 \left(\log(k)+\log(2\pi)\right)$$ Replacing, expanding the logarithms and simplifying leads to $$\log(A_n)\sim \frac 1 n\left( n (\log (27)-2)+\frac{\log (3)}{2}\right)=\log (27)-2+\frac{\log (3)}{2 n}=\log\left(\frac{27}{e^2}\right)+\frac{\log (3)}{2 n}$$ which shows the limit and how it is approached.

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