Question

using Limit Comparison Test, is the summation of 1/((n^3)(sin n)^2) from n equals 1 to infinity convergent or divergent?

Answer 1

\[\sum_{1}^{\infty} \frac{ 1 }{ n^3 \sin ^2 n}\]

Answer 2

convergent

Answer 3

divergent..

Answer 4

How did you arrive at your answers? Please help me solve.

Answer 5

use 1/n3 then u get 1/sin2n which isa positive term

hence as 1/n3 converges it tooooooooo

Answer 6

but 1/sin^2n does not converge..

Answer 7

I mean limit does not exist..

Answer 8

there is no need that the outcome of the test is converging it should be just positive

Answer 9

but limit must exist

Answer 10

we are taking n as natural that is a multiple of 57 degrees so it is positive and it is enough to be +ve for that test

Answer 11

what is the sum then..

Answer 12

test doesnt say the sum
we use it to just explain convergence of such complex series whose cant be determined

Answer 13

easily

Answer 14

https://en.wikipedia.org/wiki/Limit_comparison_test

you should get limit point such as c, but lim_>inf sin^2n does not exist..

Answer 15

Guys, accdg to the ratio test, this is inconclusive. What can you say about this? :(

Answer 16

THERE is no need that a converging series should follow all the tests 'if it satisfies any one it is enough