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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?

Lorenzo B.
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user515933
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    $$\lim_{n\to \infty}\frac{n^n}{(n+1)^n}.= \lim_{n\to \infty} \left(\frac{n+1-1}{ n+1}\right)^n = \lim_{n\to \infty} \left(1-\frac{1}{n+1}\right)^n = \dfrac{1}{e} \ne 1$$ – AgentS Mar 03 '18 at 16:51
  • Ok, but $(n+1)^n = \sum_{i=0}^{n}\binom{n}{i}n^{n-i} \iff n^n(1+...)$ where $...$ are factors in which the limit variable is at denominator so applying the limit they are zero. This identity isn't right? – user515933 Mar 03 '18 at 16:58
  • I think you're incorrect in assuming that $\binom{n}{1}\frac{1}{n}$ vanishes in the limit $$\lim_{n\to\infty} \binom{n}{1}\frac{1}{n} = 1 \ne 0$$ – AgentS Mar 03 '18 at 17:00
  • Yes, you're right! Thank you! – user515933 Mar 03 '18 at 17:02
  • Np, it seems $$ \lim_{n\to\infty} \binom{n}{k}\frac{1}{n^k} = \dfrac{1}{k!} $$ – AgentS Mar 03 '18 at 17:04
  • Hence for the denominator you should get $$ \lim \left[1+\binom{n}{1}\frac{1}{n}+ \binom{n}{1}\frac{1}{n^2}+\cdots \right]= 1 + \dfrac{1}{1!} +\dfrac{1}{2!}+\cdots = e $$ – AgentS Mar 03 '18 at 17:08

2 Answers2

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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: $$\lim_\limits{n\to\infty} \frac{2n^n}{(n+1)^n}=\frac{2}{\lim_\limits{n\to\infty} \left(1+\frac{1}{n}\right)^n}=\frac{2}{e}<1.$$

farruhota
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Hint:

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

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

lab bhattacharjee
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