Question

Determine whether ln(n/(3n+1)) diverges or converges.

Answer 1

*

Answer 2

diverges or converges where? at x=0? at x=∞?

Answer 3

as x--> infinity

Answer 4

I'm pretty sure that converges, but I forget how to explain why.

Answer 5

\[\lim_{x\to a}\ln(f(x))=\ln(\lim_{x\to a}f(x))\]

Answer 6

can you take\[\lim_{n\to\infty}\frac n{3n+1}\]?

Answer 7

Wouldn't that be indeterminate, infinity/infinity?

Answer 8

not if you divide numerator and denominator by n first

Answer 9

The +1 in the denominator doesn't make a difference at infinity, so take n/(3n).

Answer 10

^another valid way to see it

Answer 11

I sometimes forget how to run the numbers, but I can visualize my way through most anything! ;-)

Answer 12

ohh ok, I see. So that would allow me to see that the limit is 1/3. I am supposed to use a convergence test to show this though.

Answer 13

If you want to see where the limit converges to, take LN(1/3).

Answer 14

yes, this *is* the convergence test; checking if the limit converges

Answer 15

Oh, I have a list from my prof of the root test, ratio test, comparison test, p-series test, integral test, geometric series test, and test for divergence. I suppose though if I can find the limit that is proof that the series converges

Answer 16

oh you are doing a series!?

Answer 17

then the series only converges if the limit of it's terms is 0, which here it clearly is not

Answer 18

So as a series it diverges?

Answer 19

yes, think about what it means for the terms of the series to converge to a finite number
as we approach infinity this would be like adding \[\ln\frac13+\ln\frac13+\ln\frac13+\ln\frac13+....\]for infinity. this will obviously not converge as we keep adding the same number to it

Answer 20

Ah, that explains the 'n' instead of 'x'

Answer 21

if the limit of the terms of the series converge to zero the series will end up adding \[0+0+0+0+...\]as \(n\to\infty\), so the series can be finite as we don't keep adding stuff to it
make sense?

Answer 22

It just clicked. I understand now, using the nth term test for divergence I can justify my reasoning. Thanks!

Answer 23

Happy to help :)