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$\lim_{n \to \infty} (\frac{(n+1)(n+2)\dots(3n)}{n^{2n}})^{\frac{1}{n}}$ is equal to :

  1. $\frac{9}{e^2}$
  2. $3 \log3−2$
  3. $\frac{18}{e^4}$
  4. $\frac{27}{e^2}$

My attempt :

$\lim_{n \to \infty} (\frac{(n+1)(n+2)\dots(3n)}{n^{2n}})^{\frac{1}{n}}$

$=\lim_{n \to \infty} (\frac{(n+1)(n+2)\dots(n+2n)}{n^{2n}})^{\frac{1}{n}}$

$=\lim_{n \to \infty} (\frac{{n^{2n}}\{(1+1/n)(1+2/n)\dots(1+2n/n)\}}{n^{2n}})^{\frac{1}{n}}$

$=\lim_{n \to \infty} (\frac{{n^{2n}}\{(1+1/n)(1+2/n)\dots(1+2)\}}{n^{2n}})^{\frac{1}{n}}$

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

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

I'm stuck here.

Can you please explain?

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    $\begingroup$ Do you mean the limit as $n$ approaches infinity (not $x$)? The standard trick in this problem is to apply $\ln$. $\endgroup$ Apr 27, 2016 at 11:23
  • $\begingroup$ Thanks for review. I've edited typo. $\endgroup$ Apr 27, 2016 at 11:25
  • $\begingroup$ Perhaps have a look at this similar question. $\endgroup$ Apr 27, 2016 at 12:33
  • $\begingroup$ @MartinSleziak, thanks, not completely but concept is good similar. $\endgroup$ Apr 27, 2016 at 12:37

5 Answers 5

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For your reference:

Stirling's Formula is useful when one try to evaluate limits involving $n!$.

$$n!\sim\sqrt{2\pi n}\left(\frac ne\right)^n\text{ as } n\to\infty$$

Alternative solution:

Applying Stirling's Formula:

\begin{align} L&=\lim_{n\to\infty}\left(\frac{(3n)!}{n!\cdot n^{2n}}\right)^{\frac1n} \\&=\lim_{n\to\infty}\left(\frac{\sqrt{2\pi (3n)}\left(\frac {3n}e\right)^{3n}}{\sqrt{2\pi n}\left(\frac {n}e\right)^{n}\cdot n^{2n}}\right)^{\frac1n} \\&=\lim_{n\to\infty}\sqrt3^{\frac1n}\cdot\frac{(\frac{3n}e)^3}{(\frac ne)\cdot n^2} \\&=\lim_{n\to\infty}\sqrt3^{\frac1n}\cdot\frac{27}{e^2} \\&=\frac{27}{e^2} \end{align}

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  • $\begingroup$ Thanks for nice explanation. $\endgroup$ Apr 27, 2016 at 11:50
  • $\begingroup$ limit of $n!$ is $\infty$, so is $n$. So First equality doesn't make sense. Writing kinda $(f-g) \to 0$ is better. Moreover, taking limit inside of limit can be wrong. Solution way is good, btw. $\endgroup$ Apr 27, 2016 at 11:54
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$$L=\lim_{n \rightarrow \infty} ((1+\frac1n)(1+\frac2n)\dots(1+\frac{2n}{n}))^{\frac{1}{n}}$$ $$\log L=\log\lim_{n \rightarrow \infty} ((1+\frac1n)(1+\frac2n)\dots(1+\frac{2n}{n}))^{\frac{1}{n}}$$ $$\log L=\lim_{n \rightarrow \infty}\log ((1+\frac1n)(1+\frac2n)\dots(1+\frac{2n}{n}))^{\frac{1}{n}}$$ $$\log L=\lim_{n \rightarrow \infty}\frac1n\log ((1+\frac1n)(1+\frac2n)\dots(1+\frac{2n}{n}))$$ $$\log L=\lim_{n \rightarrow \infty}\frac1n(\log (1+\frac1n)+\log (1+\frac2n)+\dots+\log (1+\frac{2n}{n}))$$ $$\log L=\int_0^2\log(1+x)dx$$ $$\log L=\log(1+x)x-\int\frac{x}{x+1}dx$$ $$\log L=[\log(1+x)x-x+\log|x+1|]_0^2$$ $$\log L=2\log3-2+\log3=\log(3^2)-\log(e^2)+\log3=\log(\frac{27}{e^2})$$ $$L=\frac{27}{e^2}$$

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Similarly as in this question we can use the fact that:

If $a_n$ is a sequence of positive real numbers and the limit $\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n}$ exists, then $\lim\limits_{n\to\infty} \sqrt[n]{a_n}$ also exists and $$\lim\limits_{n\to\infty} \sqrt[n]{a_n} = \lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n}.$$

For the proof see, for example Limit of ${a_n}^{1/n}$ is equal to $\lim_{n\to\infty} a_{n+1}/a_n$ or How to show that $\lim_{n \to \infty} a_n^{1/n} = l$?


Now if we apply the above to the sequence $$a_n=\frac{(n+1)(n+2)\dots(3n)}{n^{2n}}$$ we get \begin{align*} \frac{a_{n+1}}{a_n} &= \frac{(3n+1)(3n+2)(3n+3)}{n+1} \cdot \frac{n^{2n}}{(n+1)^{2(n+1)}}\\ &= \frac{(3n+1)(3n+2)(3n+3)}{(n+1)^3} \cdot \frac{n^{2n}}{(n+1)^{2n}}\\ &= \frac{(3n+1)(3n+2)(3n+3)}{(n+1)^3} \cdot \frac{1}{\left(1+\frac1n\right)^{2n}}. \end{align*} We now see that $$\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n} = \frac{3^3}{e^2}.$$

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  • $\begingroup$ Clever use of that lemma, +1 $\endgroup$ Apr 27, 2016 at 14:46
  • $\begingroup$ Thanks for nice explanation. $\endgroup$ Apr 27, 2016 at 14:47
  • $\begingroup$ I just learned something here:) Thank you Martin. (+1) $\endgroup$ Apr 27, 2016 at 14:57
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GoodDeeds' answer is very nice (and a better answer). Here is an alternate approach using Stirling's Approximation. In particular, we need $\ln(n!)=n\ln n-n+O(\log n)$.

We begin by observing that $$ \left(\frac{(n+1)(n+2)\cdots(3n)}{n^{2n}}\right)^{1/n}=\left(\frac{(3n)!}{n!n^{2n}}\right)^{1/n}. $$

Now, we continue as above, by applying the logarithm to the limit. Namely: \begin{align*} \lim_{n\rightarrow\infty}\left(\frac{(n+1)(n+2)\cdots(3n)}{n^{2n}}\right)^{1/n}&=\lim_{n\rightarrow\infty}\left(\frac{(3n)!}{n!n^{2n}}\right)^{1/n}\\ &=\operatorname{exp}\left(\ln\left(\lim_{n\rightarrow\infty}\left(\frac{(3n)!}{n!n^{2n}}\right)^{1/n}\right)\right)\\ &=\operatorname{exp}\left(\lim_{n\rightarrow\infty}\left(\ln\left(\frac{(3n)!}{n!n^{2n}}\right)^{1/n}\right)\right)\\&=\operatorname{exp}\left(\lim_{n\rightarrow\infty}\left(\frac{1}{n}\left(\ln((3n!))-\ln(n!)-\ln(n^{2n})\right)\right)\right) \end{align*} Next, we apply Stirling's approximation from above to get that this simplifies to $$ \operatorname{exp}\left(\lim_{n\rightarrow\infty}\left(\frac{1}{n}\left(3n\ln(3n)-3n+O(\ln(3n))-n\ln(n)+n+O(\ln n)-2n\ln(n)\right)\right)\right) $$ By using the identity that $\ln(3n)=\ln(3)+\ln(n)$, we can simplify the expression (many terms cancel) above to $$ \operatorname{exp}\left(\lim_{n\rightarrow\infty}\left(\frac{1}{n}\left(3n\ln(3)-2n+O(\ln(n))\right)\right)\right) $$ Since $\lim_{n\rightarrow\infty}\frac{\ln(n)}{n}=0$, the error term vanishes and we are left with $$ e^{3\ln(3)-2}=\frac{27}{e^2}. $$

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This won't guarantee that option 4 is the correct limit, but it does rule out the other three options. We have

$${(n+1)(n+2)\cdots(3n)\over n^{2n}}=\left(1+{1\over n} \right)\cdots\left(1+{n\over n} \right)\left(1+{n+1\over n} \right)\cdots\left(1+{n+n\over n} \right)\ge1\cdots1\cdot2\cdots2=2^n$$

hence

$$\left({(n+1)(n+2)\cdots(3n)\over n^{2n}}\right)^{1/n}\ge2$$

But options 1-3 are all less than $2$.

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