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Find $\limsup_{n\to\infty}(\frac{2\cdot5\cdot8\cdot\cdots\cdot(3n-4)}{3^nn!})^{1/n}$

I've tried multiplying the nominator and the denominator by what is lacking for there to be $3n!$ in the nominator, but that landed me in an algebraic mess.

I've also tried to see if we can extract something like $const^n$ from $2\cdot5\cdot8\cdot\cdots\cdot(3n-4)$, and nothing came out except for a very thin sight of $2^{n/2}$ (I think) that I don't know how to use.

I think we are supposed to use Stirling formula somewhere in the exercise.

Thank you.

Paras Khosla
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fragileradius
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4 Answers4

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Let the given sequence in question be $a_n$ and $b_n=a_n^n$. Then $$\frac {b_{n+1}}{b_n}=\frac{3n-1}{3(n+1)}$$ which tends to $1$. Thus $a_n=\sqrt[n] {b_n} $ also tends to $1$.

Paramanand Singh
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  • Let $\lim a_n=\lim a_{n+1}=c$. Do I understand correctly that $1=\lim\frac{b_{n+1}}{b_n}=\frac{{\lim a_{n+1}}^{n+1}}{{\lim a_n}^{n}}=\frac{c^{n+1}}{c^n}=c$? Thank you. (I'm trying to understand why from the fact that the ratio in your answer tends to one follows the fact that the original sequence tends to one). – fragileradius Apr 16 '19 at 09:56
  • Use this fact to conclude that $\sqrt[n]{b_n}$ also tends to $ 1$. – Minus One-Twelfth Apr 16 '19 at 10:36
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    Most probably shortest and most elegant solution (+1) – trancelocation Apr 16 '19 at 11:39
  • @fragileradius: it's a standard theorem (a logarithmic version of Cesaro-Stolz) which says that if $a_n>0$ for all $n$ and $a_{n+1}/a_n\to L$ then $\sqrt[n] {a_n} \to L$. – Paramanand Singh Apr 16 '19 at 12:10
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Consider first that $$\prod_{i=2}^n (3i-4)=-\frac{3^n \,\Gamma \left(n-\frac{1}{3}\right)}{\Gamma \left(-\frac{1}{3}\right)}$$ then $$\frac{\prod_{i=2}^n (3i-4)} {3^n \, n! }=-\frac{\Gamma \left(n-\frac{1}{3}\right)}{\Gamma \left(-\frac{1}{3}\right) n!}$$ Take logarithms and use Stirling approximations as well as Taylor series to get $$\log\left(\frac{\prod_{i=2}^n (3i-4)} {3^n \, n! } \right)=-\frac{4}{3} \log \left({n}\right)-\log \left(-{\Gamma \left(-\frac{1}{3}\right)}\right)+\frac{2}{9 n}+O\left(\frac{1}{n^2}\right)$$ Divide by $n$ and show that the limit is just $0$. So, for the whole expression, the limit is $1$.

Claude Leibovici
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2

Squeezing can also give the limit quite quickly.

First note that

  • $\bigl(\frac{2\cdot5\cdot8\cdot\cdots\cdot(3n-4)}{3^nn!}\bigr)^{1/n} = \frac{1}{3}\left(\prod_{k=1}^{n}\frac{3k-1}{k}\right)^{\frac{1}{n}}\cdot \underbrace{\frac{1}{\sqrt[n]{3n-1}}}_{\stackrel{n\to\infty}{\longrightarrow}1}$

Now you have

$$\frac{1}{3}\left(\prod_{k=1}^{n}\frac{3k-1}{k}\right)^{\frac{1}{n}} < \frac{1}{3}\left(\prod_{k=1}^{n}\frac{3k}{k}\right)^{\frac{1}{n}} = 1$$

$$\frac{1}{3}\left(\prod_{k=1}^{n}\frac{3k-1}{k}\right)^{\frac{1}{n}} > \frac{1}{3}\left(\prod_{k=2}^n\frac{3k-3}{k}\right)^{\frac{1}{n}} = \frac{1}{\sqrt[n]{3}}\left(\prod_{k=2}^n\frac{k-1}{k}\right)^{\frac{1}{n}}= \frac{1}{\sqrt[n]{3n}}\stackrel{n\to \infty}{\longrightarrow}1$$

Both together give a limit of $1$.

trancelocation
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0

An application of Bernoulli's Inequality yields $$ \begin{align} \frac{2\cdot5\cdot8\cdot\cdots\cdot(3n-4)}{3^nn!} &=\frac13\prod_{k=2}^n\frac{3k-4}{3k}\\ &=\frac19\prod_{k=3}^n\frac{k-\frac43}{k}\\ &\ge\frac19\prod_{k=3}^n\left(\frac{k-2}{k}\right)^{2/3}\\[3pt] &=\frac19\left(\frac2{(n-1)n}\right)^{2/3}\tag1 \end{align} $$ Apply $\lim\limits_{n\to\infty}n^{1/n}=1$ to get the final limit. This follows from the Binomial Theorem: $$ \begin{align} \left(1+\sqrt{\frac2n}\right)^n &\ge1+n\sqrt{\frac2n}+\frac{n(n-1)}2\frac2n\\ &=n+\sqrt{2n}\\[6pt] &\ge n\tag2 \end{align} $$ which implies $n^{1/n}\le1+\sqrt{\frac2n}$

robjohn
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