Question

does the series 1/(n^2*3^n) converge or diverge as n goes from 1 to infinty?? what test do you use??

Answer 1

if we continue to sub larger values of n we find that the denominator heads to infinity which means this functions converges onto 0

Answer 2

thats it? you can take the limit of it? ok then thanks!

Answer 3

we can understand the behaviour of the function

Answer 4

Not really. If the limit as n goes to infinity is 0, that means there still a possibility to be convergent or divergent. An applicable test to this series is the ratio test.

Answer 5

Do you know what the ratio test is?

Answer 6

you take the limit of a(n+1) divided by a(n)..?

Answer 7

Yep as n goes to infinity. If the value of the limit is less than one, then it's absolutely convergent.

Answer 8

ok thank you so much!

Answer 9

The limit is equal to 1/3, and hence the series converges. I can do the steps if you like.

Answer 10

i got it! but i have another question..how do you find the radius of convergence? what does that even mean?

Answer 11

In power series, there is a radius of convergence. That means a domain within the radius at which the series will converge.

Answer 12

do we do the ratio test again and the limit we get is L and we just find 1/L and thats the radius of convergence??