limit with consecutive factorials (1 Viewer)

juantheron · Mar 28, 2023

Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\bigg(((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}}\bigg)$

tywebb · Mar 28, 2023

$e^{-1}$ which I got from wolframalpha.

tywebb · Mar 28, 2023

So how to do it without wolframalpha?

$\begin{aligned}\text{By the Stolz-Ces\`aro theorem }\lim_{n\rightarrow\infty}(((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}})&=\textstyle\lim\limits_{n\rightarrow\infty}\frac{((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}}}{n+1-n}\\&=\textstyle\lim\limits_{n\rightarrow\infty}\frac{(n!)^\frac{1}{n}}{n}\\&=\textstyle\frac{1}{e}\cdot\lim\limits_{n\rightarrow\infty}\left(\frac{e^nn!}{\sqrt{2\pi n}\cdot n^n}\right)^\frac{1}{n}\cdot\lim\limits_{n\rightarrow\infty}(\sqrt{2\pi n})^\frac{1}{n}\\&=\textstyle\frac{1}{e}\cdot1\cdot1\text{ by Stirling's formula}\\&=e^{-1}\end{aligned}$

juantheron · Mar 28, 2023

Thanks Tywebb.

Can we solve it without using Stolz and Stirling approximation like definite integration

If yes then please explain me.

tywebb · Mar 29, 2023

You can apply integration to the second part thus avoiding Stirling's formula but for now I'll keep the Stolz–Cesàro theorem for the first part. Maybe find another way for that later.

$\begin{aligned}\text{By the Stolz-Ces\`aro theorem }\ln\lim_{n\rightarrow\infty}(((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}})&=\ln\textstyle\lim\limits_{n\rightarrow\infty}\frac{((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}}}{n+1-n}\\&=\ln\textstyle\lim\limits_{n\rightarrow\infty}\frac{(n!)^\frac{1}{n}}{n}\\&=\textstyle\lim\limits_{n\rightarrow\infty}\frac{1}{n}\sum_{r=1}^n\ln\frac{r}{n}\\&=\textstyle\int_0^1\ln x\ \!dx\\&=-1\end{aligned}$

$\text{Hence } \lim_{n\rightarrow\infty}(((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}})=e^{-1}$

juantheron · Mar 30, 2023

Thanks Tywebb.

limit with consecutive factorials (1 Viewer)

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