Monday, 11 July 2016

calculus - Limit of nth root of a ratio that goes to infinity



Prove that
lim



(Edit: as noted in the answers the actual limit is 4; I stand corrected.)



The argument of the nth root goes to infinity since it is larger than n, also it grows faster than than n^k for any k... but taking the nth root of any of these would make the limit 1.



Is there a simple argument or should I use some trick...?



Answer



Set a_n=\binom{2n}{n}. Then if the quotient sequence \frac{a_{n+1}}{a_n} converges to a limit L, the root sequence \sqrt[n]{a_n} has the same limit.



As
\frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)}{(n+1)^2}=4·\frac{1+\frac1{2n}}{1+\frac1n}
does indeed converge, both sequences have the same limit L=4.







Bonus info: See also Newton's binomial series where
(1-x)^{-\frac12} =\sum_{k=0}^\infty\binom{-\frac12}k· (-x)^k =\sum_{k=0}^\infty\binom{2k}{k}·\left(\frac x4\right)^k
which has, as all binomial series, R=1 as radius of convergence, giving also the limit of the root sequence as 4 by the theorem of Cauchy-Hadamard on the sharpness of the root formula for that radius.


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}...