Saturday 21 January 2017

calculus - What is the limit of the sequence n!/4^n?





I am trying to find the limit of the sequence by using root test and I don't understand why the limit is not zero?
(the answer is inf).


Answer



By the root test:



$$\begin{array}{rcl}

\displaystyle \limsup_{n\to\infty} \sqrt[n]{a_n} &=& \displaystyle \limsup_{n\to\infty} \sqrt[n]{\dfrac{n!}{4^n}} \\
&=& \displaystyle \dfrac14 \limsup_{n\to\infty} \sqrt[n]{n!} \\
&=& \displaystyle \dfrac14 \limsup_{n\to\infty} \sqrt[n]{\exp\left(n \ln n - n\right)} \\
&=& \displaystyle \dfrac14 \limsup_{n\to\infty} \exp\left(\ln n - 1\right) \\
&=& \displaystyle \dfrac1{4e} \limsup_{n\to\infty} n \\
&=& \infty
\end{array}$$



Hence the sequence diverges to infinity.


No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...