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

  • 1
    $\begingroup$ Do you mean the limit as $n$ approaches infinity (not $x$)? The standard trick in this problem is to apply $\ln$. $\endgroup$ – Michael Burr Apr 27 '16 at 11:23
  • $\begingroup$ Thanks for review. I've edited typo. $\endgroup$ – Mithlesh Upadhyay Apr 27 '16 at 11:25
  • $\begingroup$ Perhaps have a look at this similar question. $\endgroup$ – Martin Sleziak Apr 27 '16 at 12:33
  • $\begingroup$ @MartinSleziak, thanks, not completely but concept is good similar. $\endgroup$ – Mithlesh Upadhyay Apr 27 '16 at 12:37

$$L=\lim_{n \rightarrow \infty} ((1+\frac1n)(1+\frac2n)\dots(1+\frac{2n}{n}))^{\frac{1}{n}}$$ $$\log L=\log\lim_{n \rightarrow \infty} ((1+\frac1n)(1+\frac2n)\dots(1+\frac{2n}{n}))^{\frac{1}{n}}$$ $$\log L=\lim_{n \rightarrow \infty}\log ((1+\frac1n)(1+\frac2n)\dots(1+\frac{2n}{n}))^{\frac{1}{n}}$$ $$\log L=\lim_{n \rightarrow \infty}\frac1n\log ((1+\frac1n)(1+\frac2n)\dots(1+\frac{2n}{n}))$$ $$\log L=\lim_{n \rightarrow \infty}\frac1n(\log (1+\frac1n)+\log (1+\frac2n)+\dots+\log (1+\frac{2n}{n}))$$ $$\log L=\int_0^2\log(1+x)dx$$ $$\log L=\log(1+x)x-\int\frac{x}{x+1}dx$$ $$\log L=[\log(1+x)x-x+\log|x+1|]_0^2$$ $$\log L=2\log3-2+\log3=\log(3^2)-\log(e^2)+\log3=\log(\frac{27}{e^2})$$ $$L=\frac{27}{e^2}$$

  • 1
    $\begingroup$ Thanks for nice explanation. $\endgroup$ – Mithlesh Upadhyay Apr 27 '16 at 11:28

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}

  • $\begingroup$ Thanks for nice explanation. $\endgroup$ – Mithlesh Upadhyay Apr 27 '16 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. $\endgroup$ – student forever Apr 27 '16 at 11:54

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 $$\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n} = \frac{3^3}{e^2}.$$

  • $\begingroup$ Clever use of that lemma, +1 $\endgroup$ – Gabriel Romon Apr 27 '16 at 14:46
  • $\begingroup$ Thanks for nice explanation. $\endgroup$ – Mithlesh Upadhyay Apr 27 '16 at 14:47
  • $\begingroup$ I just learned something here:) Thank you Martin. (+1) $\endgroup$ – Mythomorphic Apr 27 '16 at 14:57

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 $$ e^{3\ln(3)-2}=\frac{27}{e^2}. $$

  • $\begingroup$ Thanks for nice explanation. $\endgroup$ – Mithlesh Upadhyay Apr 27 '16 at 11:45

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

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


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

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

  • $\begingroup$ Thanks for nice explanation. $\endgroup$ – Mithlesh Upadhyay Oct 9 '18 at 11:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.