What is $\lim_{n\to \infty }\left(\sqrt[\leftroot{-2}\uproot{2}n+1]{(n+1)!}-\sqrt[\leftroot{-2}\uproot{2}n]{n!}\right)$?

Question

So recently a friend asked me to compute this limit:

$$\lim_{n\to \infty }\left(\sqrt[\leftroot{-2}\uproot{2}n+1]{(n+1)!}-\sqrt[\leftroot{-2}\uproot{2}n]{n!}\right)$$

Question : Does the limit exist? If yes is it finite and if yes what is its value?

How do we solve this?

Edit:

Note: I am only familiar with only basics of limit solving(upto L'Hôpital's rule) and have reasons to believe that this limit can be solved using these methods. If you could keep your answer simple that should help.

Update

Here is where I have gotten so far

$$ \lim_{n\to \infty} (n+1)! ^{1\over n+1} - (n)! ^{1\over n}$$ Can be written as $$ \lim_{n\to \infty}[1*2*3*...(n+1)]^{1\over n+1} - [1*2*3*...n] ^{1\over n}$$

$$\implies \lim_{n\to \infty} [(n+1)[{1 \over n+1}* {2 \over n+1} * {3\over n+1}...* {n+1 \over n+1}]^{1 \over n+1} - (n)[{1 \over n}* {2 \over n} * {3\over n}...* {n\over n}]^{1 \over n} ]$$

(Factoring n+1 out of first expression and n from second.)

$$\implies \lim_{n\to \infty} [(n+1) e^{{1 \over n+1} (\sum_{r=1}^{n+1}ln({r\over n+1}))} - (n) e^{{1 \over n} (\sum_{r=1}^{n}ln({r\over n}))} ]$$

From here I think second limit can be solved as a integral(limit of a sum) but I cannot solve first. How can I proceed further?

Thanks!

Answer 1

We can provide a more elementary approach, assuming the limit exists.

By Stolz-Cesaro, the discrete version of L'Hôpital's rule, we have

$$\lim_{n\to\infty}\frac{\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}}1=\lim_{n\to\infty}\frac{\sqrt[n]{n!}}n=\lim_{n\to\infty}\sqrt[n]{\frac{n!}{n^n}}$$

and by the root-to-ratio limit, we also have

$$\lim_{n\to\infty}\sqrt[n]{\frac{n!}{n^n}}=\lim_{n\to\infty}\frac{(n+1)!}{(n+1)^{n+1}}\times\frac{n^n}{n!}=\lim_{n\to\infty}\frac1{\left(1+\frac1n\right)^n}$$

which, by the limit definition of $e$, is given by

$$\lim_{n\to\infty}\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}=\frac1e$$

Answer 2

Brute force, but from the Stirling formula $$ n! = \left( {\frac{n}{e}} \right)^n \sqrt {2\pi n} \left( {1 + \mathcal{O}\!\left( {\frac{1}{n}} \right)} \right), $$ one has $$ \sqrt[n]{{n!}} = \frac{n}{e}\exp \left( {\frac{1}{2n}\log (2\pi n)} \right)\left( {1 + \mathcal{O}\!\left( {\frac{1}{{n^2 }}} \right)} \right) = \frac{n}{e} + \frac{1}{2e}\log (2\pi n) + \mathcal{O}\!\left( {\frac{{\log ^2 n}}{n}} \right). $$ This gives $$ \sqrt[{n + 1}]{{(n + 1)!}} - \sqrt[n]{{n!}} = \frac{1}{e} + \frac{1}{2e}\log \left( {\frac{{n + 1}}{n}} \right) + \mathcal{O}\!\left( {\frac{{\log ^2 n}}{n}} \right) = \frac{1}{e} + \mathcal{O}\!\left( {\frac{{\log ^2 n}}{n}} \right). $$ Thus the limit is $1/e$.

Answer 3

Calculation of the limit expressing the difference in the statement by strings whose limit is known. $$\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!} =\sqrt[n]{n!}\left(\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}-1\right) =\sqrt[n]{n!}\cdot\frac{x_{n}-1}{\ln{x_{n}}}\cdot\ln\sqrt[n(n+1)]{\frac{(n+1)^n}{n!}}= $$ $$=\frac{\sqrt[n]{n!}}{n+1}\cdot\frac{x_{n}-1}{\ln{x_{n}}} \cdot\ln\frac{n+1}{\sqrt[n]{n!}}\rightarrow \frac{1}{e}\cdot1\cdot\ln{e}=\frac{1}{e}$$

Answer 4

So I'm providing a first-principles approach, after seeing that you're unable to understand most of the answers that were posted.

We know that $$n! = n \cdot (n-1) \cdot (n-2) \cdot \cdot \cdot \cdot 2 \cdot 1$$ Taking a logarithm both sides, we get $$\ln n! = \sum_{r=1}^n\ln r$$ Looking at this summation and comparing it with the $\ln x$ function, we see that they are pretty similar, especially as $n \to \infty$ Since they are so similar, we can approximately replace the summation with a definite integral as follows: $$\ln n! \approx \int_1^n \ln x dx = n\ln n - n + 1$$

This is the principle behind the Stirling approximation, the formula that Gary used in his answer. It's a rough approximation: a more precise one uses the Euler-MacLaurin series but since our limit involves the difference of two of these terms, we can use this approximation as the error would be subtracted.

Taking an exponent and the nth root both sides gives us a very usable form of the term $(n!)^\frac 1n$ $$n! = \left( \frac ne \right)^ne \implies (n!)^\frac 1n = \frac ne \cdot e^\frac 1n$$

Now we can simply substitute the values of $n+1$ and $n$ in the obtained function and find their difference to obtain the limit. This is a fairly simple task and makes use of the taylor series expansion of $e^x$, which you should know.

$$L = \underset{n \to \infty}{\lim} \space \sqrt[\leftroot{-2}\uproot{2}n+1]{(n+1)!}-\sqrt[\leftroot{-2}\uproot{2}n]{n!} = \frac {n+1}e\cdot\sqrt[\leftroot{-2}\uproot{2}n+1]{e} - \frac {n}e\cdot\sqrt[\leftroot{-2}\uproot{2}n]{e}$$

$$ = \lim_{n \to \infty} \frac ne \left( \sqrt[\leftroot{-2}\uproot{2}n+1]{e} - \sqrt[\leftroot{-2}\uproot{2}n]{e}\right) + \frac {\sqrt[\leftroot{-2}\uproot{2}n]{e}}e$$

let $t \to 0 \implies n = \frac 1t$.

$$L = \lim_{t \to 0} \frac 1{et}\left( e^\frac{t}{t+1} - e^t \right) + \frac{e^\frac{t}{t+1}}{e}$$

Applying a taylor series expansion for $e^x$, we get

$$L = \lim_{t \to 0} \frac 1{et}\left( 1 + \frac t{t+1} + \frac {t^2}{(t+1)^2 \cdot 2!} + ... - 1 - t - \frac {t^2}{2!} - ...\right) + \frac {1 + \frac t{t+1} + \frac {t^2}{(t+1)^2 \cdot 2!} + ... }e$$ $$ = \lim_{t \to 0} \frac 1{e}\left( \frac {-t}{t+1} + O(t^2) \right) + \frac 1e$$ $$L = \frac 1e$$

which is the final answer.