Question

Use comparison test to determine convergence

The series of ((2^(n)+1)/(n2^(n)-1)) n=1 b=infinity

Answer 1

\[\sum \frac{2^n+1}{n(2^n-1)}\]?

Answer 2

the den.....n isnt outside the ( )

Answer 3

\[\sum \frac{2^n+1}{n2^n-1}\]

Answer 4

don't think it really makes any difference

Answer 5

well if u factor the n it would change to n2^n-n

Answer 6

ok remember what series i said what the archetypical example of a divergent series even though the terms go to zero?

Answer 7

(1/n)?

Answer 8

right

Answer 9

so before we do any proof or anything, lets look at what we got

Answer 10

\[\sum \frac{2^n+1}{2^n-1}\] certainly diverges, because the limit of the terms is not even 0, it is 1

Answer 11

Bn=(1/n) right? and i know what An is

Answer 12

then if we put an \(n\) in the denominator and get
\[\sum \frac{2^n+1}{n2^n-1}\] the terms do go to zero, but not fast enough because you only have the \(n\) down there

Answer 13

this is my explanation not of how to show that it diverges, but how to know before you start that it diverges, so you can proceed
now you have the job of proving it, and you are right, you can compare it to \(\frac{1}{n}\)

Answer 14

and now i am off because i am on east coast time and it is late
i am sure you can finish from there
you will get used to these pretty fast i am sure

Answer 15

thank you so much for all of your help! seriously i want a good grade on this exam and so far i feel very confident...working through these practice problems with someone is such a great help!

Answer 16

you are quite welcome
good luck with your exam, i am sure you will do well

Answer 17

i have until wednesday to study so ill be on here a lot going over problems:) thank you