Question

using limit comparison test

Answer 1

im pretty sure im going to be using the limit comparison test but his just looks divergent from the start.

Answer 2

first use the eyeball test
then use the limit comparison test to verify your answer

Answer 3

\[\sum_{n=1}^{\infty} \frac{ 4^{n+1} }{ 3^{n}-2}\]

Answer 4

looks like geometric with \(r=\frac{4}{3}\) so diverges
guess you could compare it to that one

Answer 5

clear or no?

Answer 6

yeah it seems to be divergent as well because 0<original< 4/3

Answer 7

i mean 0<4/3<original*

Answer 8

yeah actually you can use the straight up comparison test
since \(3^n-2<3^{n+1}\)you have
\[\frac{4^{n+1}}{3^n-2}>(\frac{4}{3})^{n+1}\]