((-1^n)/ n sin (n) convergent or divergent? why?

Answer 1

Let's think about what is going on.

First, I am assuming that you mean (-1)^n/(n*sin(n))

For even values of 'n' this will be positive. For odd values of 'n' this will be negative. Therefore, it only converges if 1/(n*sin(n)) goes to zero.

I believe that we can show that for any integer N, we can find a value n>N such that sin(n) < 1/n. If this is true, then n*sin(n) is less than one, so 1/(n*sin(n) is greater than one -- so the sequence does not converge to zero.

I'll leave it up to you to show that you can find sin(n) < 1/n -- but you might want to look at the Taylor's series for 'sin' to find it.

Good luck.

Answer 2

l'hopital's doesn't give a clear answer. i think it diverges. take the absolute value (to eliminate the numerator). as n increases, there will be values of sin(n) which are "extremely" close to zero, so n*sin(n) will be close to zero, so the quotient will be "large." the numerator and the sign of sine merely whip it from pos to neg but it gets "large."

Answer 3

I think i should be divergent since if u substitute numbers into n, the answer will be smaller n smaller or to infinity.

Answer 4

divergent because its negative (-1)

Answer 5

You are missing a ) somewhere so no one knows your true equation

Answer 6

You tell us