Saturday, 23 August 2014

complex numbers - How can I compute the limit of this sequence: sqrt[n]sinn?



I need to calculate the limit of the following sequence:




limnnsin(n)



where the n-th root of a negative number is defined as the principal complex root.



I suspect the answer to be 1, but I do not know how to prove it.


Answer



The problem boils down to proving that sin(n) cannot be too close to zero for small values of n.



We know that π is a trascendental number with a finite irrationality measure. In particular, the inequality

|πpq|1q10


may hold only for a finite number of rational numbers pq, hence (since |sinx|K|xkπ| when x is close to kπ, thanks to Adayah) in the general case |sin(n)| is greater than Cn9 for some constant C. That is enough to ensure that the wanted limit is 1 by squeezing.


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...