A problem of limit: $\lim\limits_{n\to \infty}\left\{\frac{(n+1)^{n}}{n!} \right\}^{(1/n)}$

Question

I need to find the limit of $$\lim\limits_{n\to \infty}\left\{\left(\frac{2}1\right) \left(\frac{3}2\right)^{2} \left(\frac{4}3\right)^{3} ... \left(\frac{n+1}n\right)^{n} \right\}^{(1/n)}$$

Cancelling out the same quantities in numerators and denominators, I have reached the step: $$\lim\limits_{n\to \infty}\left\{\frac{(n+1)^{n}}{n!} \right\}^{\frac1n}$$

Now I am stuck here. I have searched google for help, but found only the result for $\lim\limits_{n\to\infty}\left\{\frac{1}{n!} \right\}^{(\frac1n)}$.

Which formulae/properties should I use now to proceed from this stage?

Edit: The options for the answer are: a) $e$, b) $1/e$, c) $\pi$, d) $1/\pi$

See also: https://math.stackexchange.com/questions/475786/how-to-compute-lim-n-rightarrow-infty-frac1n-left-2n12n2-cdots2nn and https://math.stackexchange.com/questions/465075/find-lim-limits-n-to-infty-frac1n-sum-limits2n-r-1-fracr-sq — lab bhattacharjee, Apr 08 '18 at 05:41
Looking at your edit and the options, I bet that the limit is for $n \to \infty$ and not for $n\to 0$. One more typo in a textbook ? — Claude Leibovici, Apr 08 '18 at 09:03

score 2 · Answer 1 · answered Apr 08 '18 at 06:47

2

Taking logarithm on both the sides, we get $$\log(y_n)=\log((\frac{(n+1)^n}{n!})^{\frac{1}{n}})$$ $$=\frac{1}{n}\{\log\frac{2}{1}+2\log\frac{3}{2}+...+n\log\frac{n+1}{n}\}.$$ Using Cauchy's First theorem on limits $$\lim_{n\to\infty}\frac{1}{n}\{\log\frac{2}{1}+2\log\frac{3}{2}+...+n\log\frac{n+1}{n}\}=\lim_{n\to\infty}n\log\frac{n+1}{n}=1$$

answered Apr 08 '18 at 06:47

Am I correct? do you find anything wrong in these steps?please tell me. – Apr 08 '18 at 06:54
The question states $\lim\limits_{n\to0}$, you used infinity. – Kawrno Apr 08 '18 at 08:35
2

@Kawrno: $n\to 0$ does not make sense here as $n$ is clearly a positive integer. The right answer is $e$ as given here. – Paramanand Singh Apr 08 '18 at 09:15
1

Another source for the theorem used can be found here – B. Mehta Apr 17 '18 at 00:40
Thank you very [email protected] – Apr 17 '18 at 01:27

Claude Leibovici · Answer 2 · 2018-04-09T02:59:28.200

As you say, $n \to 0$ is rather unusual. So, let us consider the case of $x \to 0$ for

$$A=\left(\frac{(x+1)^x}{x!}\right)^{\frac{1}{x}}=\frac{x+1}{ \Gamma (x+1)^{\frac{1}{x}}}\implies \log(A)=\log(x+1)-\frac 1x\log (\Gamma (x+1))$$ Now, close to $x=0$ (have a look here), $$\log (\Gamma (x+1))=-\gamma x+\frac{\pi ^2 x^2}{12}+O\left(x^3\right)$$ $$\frac 1x\log (\Gamma (x+1))=-\gamma +\frac{\pi ^2 x}{12}+O\left(x^2\right)$$ making $$\log(A)=\gamma +\left(1-\frac{\pi ^2}{12}\right) x+O\left(x^2\right)$$ Continuing with Taylor $$A=e^{\log(A)}=e^\gamma \left(1+\left(1-\frac{\pi ^2}{12}\right) x+O\left(x^2\right) \right)$$

For illustration purposes, let $x=10^{-k}$ and compare the values $$\left( \begin{array}{ccc} k & \text{approximation}& \text{exact} \\ 0 & 2.097271488 & 2.000000000 \\ 1 & 1.812692325 & 1.811281476 \\ 2 & 1.784234409 & 1.784219608 \\ 3 & 1.781388617 & 1.781388468 \\ 4 & 1.781104038 & 1.781104036 \\ 5 & 1.781075580 & 1.781075580 \end{array} \right)$$

score 0 · Answer 3 · answered Apr 08 '18 at 05:39

0

Take the $\log$ and you find $$\log(n+1) - \log(n) + \log(n) - \frac{1}{n}\sum_{k=1}^n \log(k) = \log(1+\frac{1}{n}) - \frac{1}{n}\sum_{k=1}^n \log(\frac{k}{n}) \rightarrow -\int_0^1 \log(x) \mathrm d x = 1$$

as $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$

answered Apr 08 '18 at 05:39

Kroki

13,135

How to transform $\lim\limits_{n\to{\infty}}$ to $\lim\limits_{n\to0}$ ? – Kawrno Apr 08 '18 at 08:42
@Kawrno $n\rightarrow 0$ does not make sense if $n$ is integer. – Kroki Apr 08 '18 at 08:46
I wrote as the question was written. – Kawrno Apr 08 '18 at 12:04

Mike · Answer 4 · 2018-04-16T22:31:25.613

-1

Well, $e$ is the limit of $\left(\frac{(n+1)^n}{n!}\right)^{1/n}$ as $n$ goes to infinity. Indeed:

By Stirling's Approximation, $n!$ is $(1+o(1)\left(\frac{n}{e}\right)^n\sqrt{2\pi n}$. So $\left(\frac{(n+1)^n}{n!}\right)^{1/n}$ is

$[\sqrt{2\pi n}]^{1/n} \left(\left(\frac{n}{e(n+1)}\right)^n\right)^{1/n}$ which is $e^{-1}$.

Was that the question that was meant to be asked?

edited Apr 16 '18 at 22:31

answered Apr 16 '18 at 17:21

Mike

20,434

A problem of limit: $\lim\limits_{n\to \infty}\left\{\frac{(n+1)^{n}}{n!} \right\}^{(1/n)}$

4 Answers4