Question

use the limit comparison test to determine whether sigma 2^n/3^n-1 converges or diverges. I solved the problem but i am stuck at this point lim n-->inf 3^n/3^n-1

Answer 1

I will just continue from where you're stuck, assuming all you have done so far is right.

Answer 2

\[\lim_{n \rightarrow \infty}{3^n \over 3^n-1}=1\] (since they both grow equally).

Answer 3

i do not understand how did u got one

Answer 4

Factor out 3^n from the numerator and denominator.

Answer 5

\[\frac{3^n}{3^n-1}=\frac{3^n}{3^n(1-\frac{1}{3^n})}=\frac{1}{1-\frac{1}{3^n}}\]

Answer 6

oh okay i see it

Answer 7

The limit as n approaches infinity is \[\frac{1}{1-0}=1\]

Answer 8

Hey Lucas :)

Answer 9

Hi Anwar :)

Answer 10

i have a question in a problem like this
sigma ( 1/2n-1-1/2n) what do i compare it to

Answer 11

brb...one sec

Answer 12

k

Answer 13

Can you write that expression in the equation editor, because going off what you've written, your expression looks equal to -1.

Answer 14

?

Answer 15

\[\sum_{1}^{\infty} (1/ 2n-1-1/2n)\]

Answer 16

how do you write a fraction

Answer 17

There's a box in the equation editor, or you can type,

frac{}{}

and put your numerator stuff in the first box, denominator stuff in the second.

Answer 18

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

Answer 19

Okay...now I can look at it :)

Answer 20

:)

Answer 21

i forgot we have to use the limit comparison test here

Answer 22

I think you'd be best to combine your fractions and use the limit comparison on the result. Your summand combines as\[\frac{1}{4n^2-2n}\]

Answer 23

This is asymptotically equivalent to 1/(4n^2), which means you can eyeball the fact you could compare it to 1/n^2 for limit comparison.

Answer 24

\[\frac{\frac{1}{4n^2-2n}}{\frac{1}{n^2}}=\frac{n^2}{4n^2-2n}=\frac{1}{4-2/n} \rightarrow \frac{1}{4}\]as n goes to infinity. The limit is positive and finite, so by the limit comparison test, since Sum(1/n^2) converges, your series does too.

Answer 25

hey i got the same things i did along with u thank u for ur help

Answer 26

you're welcome