Why is this not true $\lim_{n\to \infty}\sqrt[n]{(n!)} = 1$ ? Root Test For $\frac{(n!)^2}{(2n)!}$.

Question

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.

Answer 1

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

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

Answer 2

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 $n\rightarrow\infty$.

Answer 3

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 $n\to\infty$.

Answer 4

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

Answer 5

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 $$\sqrt[n]{n!}\simeq \frac n e$$

Answer 6

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