Evaluation of $ \lim_{n\rightarrow \infty}\bigg(((n+2)!)^{\frac{1}{n+2}}-((n!))^{\frac{1}{n}}\bigg)$ using Definite integration

Question

Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\bigg(((n+2)!)^{\frac{1}{n+2}}-((n!))^{\frac{1}{n}}\bigg)$ using Definite integration

My Try : Let we assume

$\displaystyle l_1=\lim_{n\rightarrow \infty}((n+2)!)^{\frac{1}{n+2}}\Longrightarrow \ln(l_1)=\lim_{n\rightarrow \infty}\frac{1}{n+2}\ln((n+2)!)$

$\displaystyle \ln(l_1)=\lim_{n\rightarrow \infty}\frac{1}{n+2}\ln[1\cdot 2\cdot 3\cdots n\cdot (n+1)\cdot (n+2)]$

$\displaystyle \ln(l_1)=\lim_{n\rightarrow \infty}\frac{1}{n+2}\sum^{n+2}_{r=1}\ln(r)$

Similarly $\displaystyle l_2=\lim_{n\rightarrow \infty}((n!)^{\frac{1}{n}}\Longrightarrow \ln(l_2)=\lim_{n\rightarrow \infty}\frac{1}{n}\sum^n_{r=1}\ln(r)$

Help me how can i convert above limit into limit as a sum

Answer 1

Let $a_n:=((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}}$.

Then, we have $$\lim_{n\to\infty}\bigg(((n+2)!)^{\frac{1}{n+2}}-((n!))^{\frac{1}{n}}\bigg)=\lim_{n\to\infty}(a_{n+1}+a_n)$$

Now, let us first find $\displaystyle\lim_{n\to\infty}a_n$.

Letting $b_n:=\frac{((n+1)!)^{\frac{1}{n+1}}}{(n!)^{\frac 1n}}$, we can write $a_n$ as $$a_n=(n!)^{\frac 1n}(b_n-1)=\underbrace{\frac{(n!)^{\frac 1n}}{n}}_{c_n}\cdot\underbrace{\frac{b_n-1}{\ln b_n}}_{d_n}\cdot\underbrace{\ln(b_n^n)}_{f_n}$$

Since we have $$\lim_{n\to\infty}\ln(c_n)=\lim_{n\to\infty}\frac 1n\ln\bigg(\frac{n!}{n^n}\bigg)=\lim_{n\to\infty}\frac 1n\sum_{k=1}^{n}\ln\bigg(\frac{k}{n}\bigg)=\int_0^1\ln xdx=-1$$ we get $$\lim_{n\to\infty}c_n=\frac 1e$$

Also, since we have $$\lim_{n\to\infty}b_n=\lim_{n\to\infty}\frac{(n+1)c_{n+1}}{nc_n}=\lim_{n\to\infty}\frac{n+1}{n}\cdot\frac{c_{n+1}}{c_n}=1$$ we get $$\lim_{n\to\infty}d_n=\lim_{n\to\infty}\bigg(\ln\bigg(1+(b_n-1)\bigg)^{\frac{1}{b_n-1}}\bigg)^{-1}=(\ln e)^{-1}=1$$

Also, we have $$\lim_{n\to\infty}f_n=\lim_{n\to\infty}\ln\bigg(\frac{(n+1)!}{n!}\cdot\frac{1}{((n+1)!)^{\frac{1}{n+1}}}\bigg)=\lim_{n\to\infty}\ln\bigg(\frac{1}{c_{n+1}}\bigg)=1$$

Therefore, we get $$\lim_{n\to\infty}a_n=\lim_{n\to\infty}c_nd_nf_n=\frac 1e\times 1\times 1=\frac 1e$$

Since we have $\displaystyle\lim_{n\to\infty}a_{n+1}=\frac 1e$, we finally get $$\lim_{n\to\infty}\bigg(((n+2)!)^{\frac{1}{n+2}}-((n!))^{\frac{1}{n}}\bigg)=\lim_{n\to\infty}(a_{n+1}+a_n)=\color{red}{\frac{2}{e}}$$