Tuesday 16 October 2018

calculus - Proving that $n over 2^n$ converges to $0$




I'm completely clueless on this one.
I can easily calculate the limit using L'Hopital's rule, but proving that the series is converging to 0 is far more tricky.



$$a_n= {n \over 2^n}$$




Any help?


Answer



Using the fact that $2^n>n^2$ for $n > 4$, we have: $$0 \leq a_n < \frac{n}{n^2}=\frac{1}{n}.$$



Hence, $\displaystyle \lim_{n \to \infty}a_n=0.$


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}...