I need to calculate the limit of the following sequence:
limn→∞n√sin(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
|π−pq|≤1q10
may hold only for a finite number of rational numbers pq, hence (since |sinx|≥K|x−kπ| 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.
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