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This question already has an answer here:

Let $x_n \colon= \sqrt[n]{n!} $ for all $n \in \mathbb{N}$.

Then how to determine rigorously whether the sequence $\left( x_n \right)_{n \in \mathbb{N} }$ converges or diverges?

And, how to find $\lim_{n \to \infty} x_n$, rigorously?

By rigorously I mean using the same machinery as has been developed by Rudin until Chap. 3, where he discusses sequences.

My Attempt:

We note that, for all $n \in \mathbb{N}$ such that $n > 1$, we have $$ 1 \leq \left( x_n \right)^n \leq n^{n-1}, $$ and so $$ 1 \leq x_n \leq n^{ (n-1)/n } = \frac{ n }{\sqrt[n]{n} }. $$ However, although $$ \lim_{n \to \infty} \sqrt[n]{n} = 1, $$ we also have $$ \lim_{n \to \infty} n = +\infty. $$ Thus the squeeze theorem is not applicable.

Or, can we find some majorizing sequence converging to $1$?

Is this sequence monotonic?

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marked as duplicate by Hans Lundmark, Aloizio Macedo, Brian Borchers, Micah, user284331 Mar 21 '18 at 16:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Find the limit of $\sqrt[n+1]{(n+1)!}/\sqrt[n]{n!}$. That, if you don't want to just invoke Stirling's formula. $\endgroup$ – SphericalTriangle Mar 21 '18 at 13:18
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    $\begingroup$ By the elementary inequality $n!\geq\left(\frac{n}{e}\right)^n$, such limit is clearly $+\infty$. $\endgroup$ – Jack D'Aurizio Mar 21 '18 at 13:19
  • $\begingroup$ @SaaqibMahmood Please remember that you can choose an answer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – gimusi Mar 22 '18 at 21:16
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Hint: Consider the series $\sum \dfrac{1}{n!}$ and use the root test.

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You can derive it form Stirling's formula:$$\sqrt{2\pi}n^{n+\frac12}e^{-n}\leqslant n!\leqslant en^{n+\frac12}e^{-n}.$$

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Taking log plus Cesàro-Stolz: $$ \lim_{n\to\infty}\log\sqrt[n]{n!} = \lim_{n\to\infty}\frac{\log 1 + \cdots + \log n}n = \lim_{n\to\infty}\log n = \infty, $$ so $$ \lim_{n\to\infty}\sqrt[n]{n!} = \infty. $$

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By ratio-root criteria

$$a_n = \sqrt[n]{n!} \quad b_n=n!$$

$$\frac{b_{n+1}}{b_n} \rightarrow L\implies a_n=b_n^{\frac{1}{n}} \rightarrow L$$

thus since

$$\frac{b_{n+1}}{b_n}=\frac{{(n+1)!}}{n!}=n+1\to+\infty \implies a_n = \sqrt[n]{n!}\to +\infty $$

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From

$$(2n-k)(n-k)\gt(2n-2k)(n-k)=2(n-k)^2$$

for $0\lt k\lt n$, we have

$$\begin{align} x_{2n} &=\sqrt[2n]{(2n)!}\\&=\sqrt[2n]{(2n)(n)(2n-1)(n-1)\cdots(n+2)(2)(n+1)(1)}\\ &\gt\sqrt[2n]{2(n)^22(n-1)^2\cdots2(2)^22(1)^2}\\ &=\sqrt2\sqrt[n]{n!} \end{align}$$

for $n\ge2$. (The inequality is not strict for $n=1$, because there are no $k$'s between $0$ and $1$.) Thus the sequence has subsequences, at least, that diverge to infinity, for example

$$x_{2^{n+1}}\gt(\sqrt2)^n\to\infty$$

To show that the sequence as a whole diverges to infinity, it suffices to show that $x_{n+1}\gt x_n$, i.e.,

$$((n+1)!)^n\gt(n!)^{n+1}$$

which is to say,

$$(n+1)^n\gt n!$$

which is obviously true.

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By elementary means:

Consider the sequence

$$2^0,2^1,2^1,2^2,2^2,2^2,2^2,2^3,2^3,2^3,2^3,2^3,2^3,2^3,2^3,\cdots2^b,2^b,2^b,\cdots 2^b$$

term-wise inferior to the naturals.

We obviously have

$$n!\ge 2^{b2^b}=2^{b(n+1)/2}$$

hence the series diverges.

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