Question

Convergence / Divergence of Sequences

Answer 1

State if the sequence converges, and if it does, find the limit:

\[\ln (\frac{ 8n }{ n+1 })\]

Answer 2

@jim_thompson5910 @mww @agent0smith

Answer 3

Looks like it does not converge.

\[\large \lim_{n \rightarrow \infty } \ln \left(\frac{ 8n }{ n+1 } \right)=\ln \left( \lim_{n \rightarrow \infty } \frac{ 8n }{ n+1 } \right)= \] \[\large \ln \left( \lim_{n \rightarrow \infty } \frac{ 8n }{ n+1 } \right) = \ln \left( \lim_{n \rightarrow \infty } \frac{ 8 }{ 1 + \frac{ 1 }{ n } } \right) = \ln 8\]

I multiplied everything by 1/n in the second last step.

Answer 4

Doesn't converge since the terms never approach zero.

Answer 5

why does this not converge?

Answer 6

@agent0smith I think it does converge.
Convergence just requires the terms to go down to a finite value based on the limit

Answer 7

Maybe he is thinking the convergence test for sums

Answer 8

the sequence of terms converges to ln 8
if it were a series (a running sum of terms), then it would not converge

Answer 9

Why does it converge to ln(8)?

Answer 10

@ganeshie8 @TheSmartOne @MARC_D

Answer 11

Can you tell me what the expression \(\dfrac{8n}{n+1}\) approach to as you make \(n\) large ?

Answer 12

infinity

Answer 13

oh wait.
the exponents are the same so it would be 8 wouldnt it

Answer 14

maybe I need to brush up on my limits

Answer 15

Yes. Next we use the a crucial fact about limits of "continuous" functions :

\[\lim f(g(x)) = f(\lim g(x))\]

Answer 16

so I can write the limit inside the natural log is what your saying, which would make it converge to ln(8)?

Answer 17

Exactly!

Answer 18

Gotcha, thank you ganeshie et al