# limit with consecutive factorials (1 Viewer)

#### juantheron

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

#### tywebb

##### dangerman
$\bg_white e^{-1}$ which I got from wolframalpha.

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#### tywebb

##### dangerman
So how to do it without wolframalpha?

\bg_white \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

##### Active Member
Thanks Tywebb.

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

If yes then please explain me.

#### tywebb

##### dangerman
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.

\bg_white \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}

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

Thanks Tywebb.