Solve $\lim_{n\to +\infty}\frac{n!2^n}{n^n}$

Question

I tried to solve this limit with the ratio criteria: $$\lim_{n\to +\infty} \frac{a_{n+1}}{a_n},\text{ then }\lim_{n\to +\infty}\frac{(n+1)!\space2^{n+1}}{(n+1)^{n+1}}\frac{n^n}{n!\space2^n} \iff \lim_{x\to +\infty}\frac{2\space n^n}{(n+1)^n}.$$ I know that $(n+1)^n =\displaystyle \sum_{i=0}^{n}\binom{n}{i}n^{n-i}$.

So I can write this sum as $n^n(1+\binom{n}{1}\frac{1}{n}+.....)$. So I obtain $\displaystyle \lim_{n\to +\infty} \frac{2\space n^n}{n^n(1+\binom{n}{1}\frac{1}{n}+.....)} = 2$, and the limit is $+\infty$. But when I checked the result on Wolfram I realised that is wrong.

Can somebody help me?

Answer 1

Hint:

$$\dfrac{2^{n+1}(n+1)! n^n}{(n+1)^{n+1}2^n n!} =\dfrac2{\left(1+\dfrac1n\right)^n}$$

Use About $\lim \left(1+\frac {x}{n}\right)^n$

Answer 2

It seems you are mixing up concepts. Do you want to calculate the limit: $$\lim_\limits{n\to\infty} \frac{n!2^n}{n^n}=0$$ or check the convergence of the series (using ratio test): $$\sum_{n=1}^{\infty} \frac{n!2^n}{n^n}?$$ Note: $$\lim_\limits{n\to\infty} \frac{2n^n}{(n+1)^n}=\frac{2}{\lim_\limits{n\to\infty} \left(1+\frac{1}{n}\right)^n}=\frac{2}{e}<1.$$