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I am trying to find the limit of the sequence by using root test and I don't understand why the limit is not zero? (the answer is inf).

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    $\begingroup$ why do you think the answer should be zero? can you show us your attempt? $\endgroup$
    – DHMO
    Apr 16, 2017 at 9:39
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    $\begingroup$ $\displaystyle \lim_{n\to\infty} \dfrac{n!}{4^n} = \dfrac14 \cdot \dfrac24 \cdot \dfrac34 \cdot \dfrac44 \cdot \dfrac54 \cdots = \infty$ $\endgroup$
    – DHMO
    Apr 16, 2017 at 9:40
  • $\begingroup$ Note that $\displaystyle \limsup_{n\to\infty} \sqrt[n]{n!} = \infty$ $\endgroup$
    – DHMO
    Apr 16, 2017 at 9:42
  • $\begingroup$ See also Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0$, $x \in \Bbb R$. and other questions linked there. $\endgroup$ Apr 16, 2017 at 15:11
  • $\begingroup$ why lim sup =inf ?? $\endgroup$
    – Owen
    Apr 16, 2017 at 15:42

6 Answers 6

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A fancy and nice (imo) way to do it. Take

$$a_n:=\frac{4^n}{n!}\implies\frac{a_{n+1}}{a_n}=\frac{4^{n+1}}{(n+1)!}\frac{n!}{4^n}=\frac4{n+1}\xrightarrow[n\to\infty]{}0<1\implies \sum_{n=1}^\infty a_n$$

$$\text{converges}\;\implies \lim_{n\to\infty}a_n=0\implies\;\lim_{n\to\infty}\frac1{a_n}=\infty$$

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  • $\begingroup$ The first expression you wrote for $a_n$ is incorrect. $\endgroup$
    – Hoc Ngo
    Apr 16, 2017 at 10:04
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    $\begingroup$ I like this approach more than any other.+1 $\endgroup$
    – Paramanand Singh
    Apr 16, 2017 at 10:05
  • $\begingroup$ @HocNgo I don't understand: what is wrong? $\endgroup$
    – DonAntonio
    Apr 16, 2017 at 11:07
  • $\begingroup$ @DonAntonio, you swapped the numerator and denominator. The correct expression for $a_n$ should be $$ a_n = \frac{n!}{4^n}.$$Hmm, I just realized your $a_n$ is not the same as the one I thought in the original problem. My apology. $\endgroup$
    – Hoc Ngo
    Apr 16, 2017 at 11:13
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    $\begingroup$ @HocNgo At the end, he considers $1/a_n$, which is your limit. $\endgroup$ Apr 16, 2017 at 11:23
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One may note that

$$n!=1\times2\times3\times4\times5\times\dots\times n>24\times5^{n-4}$$

Thus,

$$a_n>\frac{24}{5^4}\times\left(\frac54\right)^n\to\infty$$

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All the answers given so far use valid techniques for showing that the sequence goes to infinity. But don't forget you can always go back to the definition of a limit. The limit of a sequence is infinite if you can choose any value, call it $K$, and find an index $k$ such that $a_n > K$ for all $n > k$. So, we can pick a value -- one billion -- and find a $k$ such that every $a_n$ past that is greater than a billion.

Can you prove that? Hints: can you show that:

  • once you are past the first few elements, the sequence is greater than one
  • once you are past the first few elements, the sequence is monotone increasing
  • that ${5 K}/{4} > K$
  • that these three facts can be combined to show a proof of the desired property
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Let $\displaystyle{u_n=\frac{n!}{4^n}}$. We observe that :

$$\frac{u_{n+1}}{u_n}=\frac{n+1}{4}\underset{n\to\infty}{\longrightarrow}+\infty$$

Hence, for $n\ge N$, with $N$ large enough :

$$\frac{u_{n+1}}{u_n}\ge2$$

and thus :

$$\forall n\ge N,\,u_n\ge 2^{n-N}u_N$$

This proves that $\displaystyle{\lim_{n\to\infty}u_n=+\infty}$

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You can use Stirling's approximation for very large $n$ , $$n! \approx n^ne^{-n} \sqrt{2\pi n}$$

Your limit becomes : $$\displaystyle \lim_{n\to\infty} \dfrac{n!}{4^n} = \lim_{n \to \infty} \Big(\frac{n}{4e}\Big)^n\sqrt{2\pi n}=\infty $$

Since $n$ is very large, doesn't matter if you divide it by $4e$, it will also be very large than $1$.

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By the root test:

$$\begin{array}{rcl} \displaystyle \limsup_{n\to\infty} \sqrt[n]{a_n} &=& \displaystyle \limsup_{n\to\infty} \sqrt[n]{\dfrac{n!}{4^n}} \\ &=& \displaystyle \dfrac14 \limsup_{n\to\infty} \sqrt[n]{n!} \\ &=& \displaystyle \dfrac14 \limsup_{n\to\infty} \sqrt[n]{\exp\left(n \ln n - n\right)} \\ &=& \displaystyle \dfrac14 \limsup_{n\to\infty} \exp\left(\ln n - 1\right) \\ &=& \displaystyle \dfrac1{4e} \limsup_{n\to\infty} n \\ &=& \infty \end{array}$$

Hence the sequence diverges to infinity.

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