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I have to conclude if the following problem is convergent or not using the root test or ratio test for positive series.

$$\frac{(n!)^2}{(2n)!}$$

I've shown it to be convergent using the ratio test, but I've run into a problem when using the root test. I've concluded that $\sqrt[n]{(n!)} \to 1$ but Wolfram Alpha claims that $\sqrt[n]{(n!)} \to \infty$.

Here is my work:

I started by reducing the expression:

$$\frac{(n!)^2}{(2n)!} = \frac{(n!)(n!)}{(2n)!}.$$ We have that $$(n!) = (n)(n-1)(n-2)\cdot...\cdot 1 $$ $$(2n)! = (2n)(2n-1)\cdot...\cdot 1 $$ $$ = 2(n)(n-1/2)(n-1)(n-3/2)(n-2)\cdot...\cdot 1 $$ we rearrange this $$ 2(n)(n-1)(n-2)(n-1/2)(n-3/2)\cdot ...\cdot 1 .$$ It should be clear that this "contains" $(n!).$ So now we have:

$$\frac{(n!)(n!)}{(2n)!} = \frac{(n!)}{(2n-1)(2n-3)\cdot ...\cdot 1 }. $$ I now applied the root test to this:

$$\sqrt[n]{\frac{(n!)}{(2n-1)(2-3)\cdot ...\cdot 1}} = {\frac{\sqrt[n]{(n!)}}{\sqrt[n]{(2n-1)(2-3)\cdot ...\cdot 1}}}. $$

I examined $\sqrt[n]{(n)}$:

$$\sqrt[n]{(n)} = e^{(ln(n)\cdot(1/n))}. $$ $e^x$ is continuous so we examine

$$\lim_{n\to \infty}(\frac{ln(n)}{n}) \to \frac{\infty}{\infty}. $$ We use L'hopital: $$\lim_{n\to \infty}(\frac{ln(n)'}{n'}) = \lim_{n\to \infty}(\frac{1/n}{1}) = \lim_{n\to \infty}1/n \to 0.$$

So we have $$\lim_{n\to \infty}\sqrt[n]{(n!)} = e^0 = 1.$$ We will have the same result given $\sqrt[n]{(n-k)}.$

We can now return to $\sqrt[n]{(n!)}$:

$$\sqrt[n]{(n!)} = \sqrt[n]{(n)}\sqrt[n]{(n-1)}\sqrt[n]{(n-2)}*..*1$$

$$\lim_{n\to \infty}\sqrt[n]{(n!)} = 1\cdot1\cdot1\cdot ... \cdot 1 = 1 .$$ I end up concluding the same thing regarding $$\lim_{n\to \infty}\sqrt[n]{{(2n-1)(2-3)\cdot ... \cdot 1}}.$$

I can't figure out why $\sqrt[n]{(n!)} \to \infty,$ according to Wolfram Alpha. Is it because it becomes an indeterminant form of $1^\infty$ ? Or have I simply made a mistake so that it never becomes $1\cdot 1\cdot 1\cdot ... \cdot 1,$ repeating infinitely?

I've looked at some of the other answers concerning the question of why $\sqrt[n]{(n!)} \to \infty$ it doesn't answer why my specific argument is incorrect, and most of them seem to be using a "trick". Given what is in the book I'm using, these seem needlessly complicated or foreign to the reader. I am simply doing this for fun, the problem is already solved by using the ratio test, so perhaps I am simply underestimating the complexity of this problem.

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    $\begingroup$ The line before "i end up concluding..." is incorrect: you can't apply arithmetic of limits to an expression in which the number of factors depends on $\;n\;$ itself! Only to an expression when the number of factors is a constant. $\endgroup$
    – DonAntonio
    Jan 20 '19 at 22:31
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    $\begingroup$ $\lim_{n\to \infty} \sqrt[n]{n} = 1$. Why would you think $\lim_{n\to \infty} \sqrt[n]{n!} = 1,$ too? $\endgroup$ Jan 20 '19 at 22:32
  • $\begingroup$ Also, please write well on this website, and not as you would writing a text message to a friend. That is, there is no word "i". There is a word "I". Also, the use of punctuation is immensely helpful, else your post comes across as one huge run-on sentence. Be respectful to other users using this site, particularly if you want to be understood, by taking care in what and how you write your questions. $\endgroup$ Jan 20 '19 at 22:38
  • $\begingroup$ The comment by @DonAntonio is the only one that answers what's actually wrong with the argument given in the question. All the answers given so far are basically repetition of what's already in this old question from 2012: math.stackexchange.com/questions/136626/… $\endgroup$ Jan 21 '19 at 7:46
  • $\begingroup$ @jordan_glen I took an extensive amount of time to try and write this as well as I could, I did attempt to capitilize i every time, but i guess i missed it? I was taught that you should never punctuate math, I don't see how it could ever be misunderstood without punctuation since there are clear breaks in the text or calculations. Why do you attribute this to malice and not incompetence? I tried my best and will follow your directions to the best of my ability in the future. $\endgroup$
    – Dimajo
    Jan 21 '19 at 12:19
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In the list $1,2,3,\dots,n,$ the number of terms greater than $n/2$ is at least $n/2.$ Thus $n!>(n/2)^{n/2}.$ It follows that

$$(n!)^{1/n} > [(n/2)^{n/2}]^{1/n} = (n/2)^{1/2} \to \infty.$$

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Let $a_{n}=\sqrt[n]{n!}$ and $b_{n}=\ln a_{n}$. Then $b_{n}=\frac{1}{n}\sum_{k=1}^{n}\ln k$. For $k\geq2$ and $x\in[k-1,k]$, we have $\ln k\geq\ln x$, so $\ln k=\int_{k-1}^{k}\ln k\,dx\geq\int_{k-1}^{k}\ln x\,dx$. Therefore \begin{eqnarray*} b_{n} & \geq & \frac{1}{n}\sum_{k=2}^{n}\ln k\\ & \geq & \frac{1}{n}\sum_{k=2}^{n}\int_{k-1}^{k}\ln x\,dx\\ & = & \frac{1}{n}\int_{1}^{n}\ln x\,dx\\ & = & \frac{1}{n}\left(n\ln n-n-1\right)\\ & = & \ln n-\frac{n+1}{n}\\ & \rightarrow & \infty \end{eqnarray*} as $n\rightarrow\infty$. Therefore $a_{n}=\exp(b_{n})\rightarrow\infty$ as $n\rightarrow\infty$.

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Per Stirling's formula that states $$ n! \sim {n^n e^{-n}\over \sqrt{2\pi n}},$$ you have $$\root{n}\of{n!} \sim \left({n^n e^{-n}\sqrt{2\pi n}}\right)^{1/n} \sim {n\over e} $$ as $n\to\infty$.

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    $\begingroup$ The solution is not self-contained because what has been assumed (i.e., Stirling's formula) is far more complicated and involving than what has been asked to prove (i.e. $\sqrt[n]{n!}\rightarrow\infty$). $\endgroup$ Jan 20 '19 at 22:51
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    $\begingroup$ If OP knew Stirling's formula, he would be mathematically mature enough and should not get stuck in that simple question. $\endgroup$ Jan 20 '19 at 23:00
  • $\begingroup$ The $\sqrt{2\pi n}$ belongs in the numerator, not the denominator. $\endgroup$ Jan 21 '19 at 0:12
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You may consider the reciprocals $\boxed{\frac{1}{\sqrt[n]{n!}}}$ and proceed as follows using GM-HM (inequality between geometric and harmonic mean) and using

  • $\sum_{k=1}^n\frac{1}{k} < \ln n +1$ for $n >1$
  • $\frac{\ln n}{n } \stackrel{n \to \infty}{\longrightarrow} 0$

\begin{eqnarray*} \frac{1}{\sqrt[n]{n!} } & \stackrel{GM-HM}{\leq} & \frac{\sum_{k=1}^n\frac{1}{k}}{n}\\ & \stackrel{\sum_{k=1}^n\frac{1}{k}< \ln n + 1}{\leq} & \frac{\ln n +1}{n}\\ & \stackrel{n \to \infty}{\longrightarrow} & 0\\ \end{eqnarray*}

It follows $\boxed{\sqrt[n]{n!} \stackrel{n \to \infty}{\longrightarrow} \infty}$

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Start with $$a_n=\sqrt[n]{n!} \implies \log(a_n)=\frac 1n \log(n!)$$ Now, by Stirling approximation $$\log(n!)=n (\log (n)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left({n}\right)\right)+\frac{1}{12 n}+O\left(\frac{1}{n^3}\right)$$ $$\log(a_n)=\log (n)-1+\frac{1}{2n} \left(\log (2 \pi )+\log \left({n}\right)\right)+O\left(\frac{1}{n^2}\right)$$ So, by the end $$\sqrt[n]{n!}\simeq \frac n e$$

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This is not rigorous, but it gives an intuitive solution (and this can be made rigorous). From Calculus we know that:

$$ \log(n!) = \sum_{k=1}^{n} \ln(k) \approx \int_1^n \ln(x) dx = n \ln n - n + C. $$

Therefore, $$ n! = e^{\log(n!)} \approx e^{n \ln n - n + C} \approx e^{n \ln n}e^{-n} = \left( \frac{n}{e}\right)^n. $$

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  • $\begingroup$ I should note that this essentially the same as Danny Pak-Keung Chen's solution (or at least it's the same idea), though his solution is more rigorous and thus less intuitive. $\endgroup$
    – JavaMan
    Jan 21 '19 at 4:19

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