Question

use the alternating series test (if possible) to determine whether the series converges or diverges

Answer 1

\[\sum_{n=1}^{\infty} \frac{ (-1)^n * [\sin(3n)]^2 }{ n }\]

Answer 2

i know an is equal to \[\frac{ [\sin(3n)]^2 }{ n }\]

Answer 3

but what do i do next

Answer 4

You have to show that this new sequence \(a_n\) converges to 0, and that this sequence is decreasing.

To do so, recall that the sine function is bounded: \(-1\le\sin x\le1\), so \(-1\le\sin3n\le1\). Squaring tells you that \(\sin^2(3n)\le1\).

Answer 5

***\(0\le\sin^2(3n)\le1\)

Answer 6

what about the n in the denominator and do i replace n+1 with n? @SithsAndGiggles