Prove: $\lim\limits _{n\rightarrow\infty} \left( ((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}} \right)=\frac{1}{e}$

Asked Jul 01 '19 at 20:47

Active Jul 01 '19 at 21:51

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I see this solution in a book but I don't understand it. Which theorems were used?

I know how I solve this limit but this solution in book as following:

$$\lim\limits_{n \rightarrow \infty} \left( (n+1) \left(\frac{(n+1)!}{(n+1)^{n+1}} \right)^{\frac{1}{n+1}}-n \left(\frac{n!}{n^n} \right)^{\frac{1}{n}} \right)$$

$$=\lim\limits_{n\rightarrow\infty}\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}}$$

$$=\lim\limits_{n\rightarrow\infty}\left( \frac{n}{n+1} \right)^n = \frac{1}{e}$$

But I don't understand which theorem use it here in the first step? Is it Alembert?

edited Jul 01 '19 at 21:18

asked Jul 01 '19 at 20:47

Kînan Jœd

You should need Stirling formula. – hamam_Abdallah Jul 01 '19 at 21:09
I tell I know how I solve it , I need understand this method . – Kînan Jœd Jul 01 '19 at 21:13
PS: This is known as a Lalescu sequence, named after the Romanian mathematician Traian Lalescu. – Alexdanut Jul 01 '19 at 21:38
Answered here – user0410 Jul 01 '19 at 21:41

Prove: $\lim\limits _{n\rightarrow\infty} \left( ((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}} \right)=\frac{1}{e}$

0 Answers0