What will be this limit? I'm having a hard time finding this....

Question

What will be this limit?

\begin{equation} \lim_{\ n\to\infty}\ \left(\frac{\left(n!\right)}{n}\right)^{\frac{1}{n}}=P\ \end{equation}

I tried it like this: its in the form (infinity)^(0), so taking natural log on both sides, we can convert it to (inf/inf) or (0/0) form, so that we can apply L'Hôpital's rule. Eventually it ended up like: \begin{equation} \lim_{n\to\infty}\ \left(\frac{\log_{e}\left(\int_{0}^{\infty}e^{-t}t^{\left(n-1\right)}dt\right)}{n}\right)\ =\ln\left(P\right) \end{equation}

So it's in the form (inf/inf) and we can apply L'Hôpital's rule. It results in:

\begin{equation} \lim_{n\to\infty}\left(\text{Digamma}\left(n\right)\right)=\ \ln\left(P\right),\text{ which tends to} +\infty \end{equation}

So it will tend to (+infinity)

But in my book, the limit is given as (1/e) I don't know how it is 1/e

Answer 1

$$\left(\frac{\left(n!\right)}{n}\right)^{\frac{1}{n}}=\left((n-1)!\right)^{\frac{1}{n}}$$

$$ \geq \sqrt[n]{\left((n-1)\times 1\right) \times \left((n-2)\times 2\right) \times \cdots \times \left( \left(n-\left\lceil \frac{n}{2} \right\rceil + 1 \right)\times \left( \left\lceil \frac{n}{2} \right\rceil - 1 \right) \right) } $$

$$ \geq \left( n-1 \right) ^ {\frac{\left\lfloor \frac{n}{2} \right\rfloor-1}{n}} \geq \left( n-1 \right)^{1/3} \text{ for } n\geq \ 17.$$

Therefore,

$$\left(\frac{\left(n!\right)}{n}\right)^{\frac{1}{n}} \to\infty.$$

Answer 2

Note that we have

$$n!\ge \left(\frac{n}{2}\right)^{n/2}$$

Therefore,

$$\left(\frac{n!}{n}\right)^{1/n}\ge \frac{\sqrt{n/2}}{n^{1/n}}$$

Can you finish now?

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Now, let's examine the limit

$$\lim_{n\to \infty}\frac{\left(n!\right)^{1/n}}{n}$$

Applying Stirling's Approximation we find that as $n\to \infty$

$$\frac{(n!)^{1/n}}{n}\sim \frac{(2\pi n)^{1/2n}(n/e)}{n}\to \frac1e$$

And we are done.

Answer 3

As Yiogoros S. Smyrlis said, i assume you mean $$\lim_{n \to \infty} \frac{(n!)^{\frac{1}{n}}}{n}$$

So, first, set it equal to a variable, say it will be x. $$x=\lim_{n \to \infty} \frac{(n!)^{\frac{1}{n}}}{n}$$ $$\log(x) = \frac{1}{n}\log n! - n\log n$$ $$=\frac{1}{n} \sum_{q=1} ^{n} \log\bigg(\frac{q}{n}\bigg)$$ Where the sum is a Riemann sum for $\int_{0} ^{1}\log xdx$, can you work it out from here?

Answer 4

Try and use Sterling's approximation which is: $$n!\sim\sqrt{2\pi n}\left(\frac ne\right)^n$$ so you get: $$\left(\frac{n!}n\right)^{1/n}\sim\left(\frac{\sqrt{2\pi n}\left(\frac ne\right)^n}n\right)^{1/n}=\frac ne\left(\sqrt{\frac{2\pi}{n}}\right)^{1/n}=\frac{n^{1-\frac1{2n}}}{e}(2\pi)^{\frac1{2n}}$$ which you can see as you increase $n$ is asymptotic to: $$\frac ne$$ which diverges. This is the same result that many calculators came to so I believe there is a mistake in the book