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(1 point) For each of the sequences below, enter either diverges if the sequence diverges, or the limit of the sequence if the sequence converges as n→[infinity]. (Note that to avoid this becoming a "multiple guess" problem you will not see partial correct answers.) A. sinn3n :

Respuesta :

eve24great2001

Answer:

limit n→[infinity], sin(3n) neither converges nor diverges.

Detailed explanation below

Step-by-step explanation:

The trigonometric function/sequence sin(3n) continues to assume a periodic values as the values of n continue to increase. That is, n→[infinity]. sin(3n) neither diverges nor converges to a limit, uniformly. Rather, the sequence continue to diverge and converge repeatedly as n→[infinity].

This is because sin(3n) will form a periodic and oscillatory sequence as n tends to infinity.

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