Let $A_n=\frac{(n+1)+(n+2)+(n+3)+.....+(n+n)}{n}$,$B_n=[(n+1)(n+2)(n+3)....(n+n)]^{1/n}$.If $\lim_{n\to\infty}\frac{A_n}{B_n}=\frac{ae}{b}$

Question

For positive integers $n$,let $A_n=\frac{(n+1)+(n+2)+(n+3)+.....+(n+n)}{n}$,$B_n=[(n+1)(n+2)(n+3)....(n+n)]^{1/n}$.If $\lim_{n\to\infty}\frac{A_n}{B_n}=\frac{ae}{b}$ where $a,b\in \mathbb{N}$ and relatively prime. Find the value of $(a+b)$.

My try:
$$\lim_{n\to\infty}\frac{A_n}{B_n}=\lim_{n\to\infty}\frac{\frac{(n+1)+(n+2)+(n+3)+.....+(n+n)}{n}}{[(n+1)(n+2)(n+3)....(n+n)]^{1/n}}$$

But I could not solve and simplify it further.

Answer 1

$A_n=\frac{3n+1}{2}$, hence $\lim_{n\to +\infty}\frac{A_n}{n}=\frac{3}{2}$. On the other hand,

$$\log B_n = \frac{1}{n}\sum_{k=1}^{n}\log(n+k) = \log n+\frac{1}{n}\sum_{k=1}^{n}\log\left(1+\frac{k}{n}\right)\tag{1}$$ but: $$\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\log\left(1+\frac{k}{n}\right)=\int_{1}^{2}\log(x)\,dx = \log 4-1,\tag{2} $$ hence $B_n$ behaves like $\frac{4n}{e}$ and: $$ \lim_{n\to +\infty}\frac{A_n}{B_n} = \color{red}{\frac{3e}{8}}.\tag{3}$$

Answer 2

You need to use Stirling's approximation to estimate $B_n$. Estimating $A_n$ is easy, since you can compute it exactly: $A_n \approx (3/2)n$. As for $B_n$, Stirling's approximation gives $$ B_n = \left[\frac{(2n)!}{n!}\right]^{1/n} \approx \frac{(2n/e)^2}{n/e} = 4n/e. $$ So $A_n/B_n \to (3/2)(e/4) = (3/8)e$.

Answer 3

There is no problem with $$A_n=\frac{3n+1}{2}$$ Now, consider that $$(B_n)^n=\frac{(2n)!}{n!}$$ and use Stirling approximation for the factorial. This leads to $$(B_n)^n\approx 2^{2 n+\frac{1}{2}} e^{-n} n^n$$ So $$B_n\approx 2^{2+\frac 1{2n}}\frac n e$$ So , for large values of $n$, $B_n\approx \frac{4n}{e}$ and $$\frac{A_n}{B_n}\approx \frac{3n+1}{2}\frac e {4n}=\frac{3n+1}{8n}e$$.

I am sure that you can take from here.