Question

sumation of n=1 going to infinity of ln (n/3n+1). Determine whether the series is convergent or divergent.

Answer 1

\[\sum_{n=1}^{\infty} \ln (n/3n+1)\]

Answer 2

This...
\[\Large \sum_{n=1}^\infty \ln\left(\frac{n}{3n+1}\right)\]
This?

Answer 3

yes

Answer 4

Okay :)
You know, the first thing you should check when you're testing convergence of series... is the underlying sequence, namely...
\[\Large \ln\left(\frac{n}{3n+1}\right)\]

What is ...
\[\LARGE \lim_{n\rightarrow \infty}\quad\ln\left(\frac{n}{3n+1}\right)= \color{red}?\]

Answer 5

1/3?

Answer 6

\[\ln (1/3)\]

Answer 7

That's right :)
And you know that for a series to be convergent, its underlying sequence must ALWAYS converge to ZERO right?

This one doesn't... therefore...?
;)

Answer 8

it is divergent.....but what test would you use to prove it? integral test? comparison test?

Answer 9

Divergence test... basic :)
The sequence does not converge to zero... that alone means the infinite series is divergent :)

Answer 10

Your instructor might call it the "Limit Test"
(Not the Limit COMPARISON test, ok?)

Answer 11

oh that's right!!

Answer 12

oh, you got it... great :D
Nice job, by the way ;)

Answer 13

haha thanks...ok so when would i know when to use this test?

Answer 14

It's the easiest test to implement, but also possibly the least useful... I think your instructor is almost always going to give you a series whose underlying sequence does, in fact, converge to zero... btw, if the sequence converges to zero, it does NOT mean the series is convergent...


It's easy to implement, anyway, so I suggest you do this limit test first... if possible, and who knows, you might get an instant point :)

Answer 15

well if it converges to 0 it is inconclusive so i would have to use another test right?

Answer 16

Yes. That's right :)

Answer 17

ok got it thanks!

Answer 18

No problem :)

Answer 19

do you think you could help me out with the limit comparison test please?

Answer 20

Sure :)
You got a sample question?

Answer 21

\[\sum_{n=1}^{\infty} (n ^{2} -5n)/ (n ^{3}+n+1)\]

Answer 22

Interesting :)
Okay, to use the Limit-Comparison test, much like with Direct Comparison, you need an intuition... what do you suspect this series to be... convergent, or divergent?

Answer 23

well i would compare it to 1/n and that diverges, so divergent?

Answer 24

That's right :)
So, you know the gist of the Limit-Comparison test?

Answer 25

You put both sequences, one on top of the other (LOL) and get the limit as n goes to infinity...
\[\LARGE \lim_{n\rightarrow \infty}\frac{a_n}{b_n}\]
If it goes to some number greater than 0, then either both converge or both diverge.

Answer 26

If it goes to 0, then if a_n converges, then so does b_n
If it goes to infinity, then if b_n converges, then so does a_n

Answer 27

if it goes to some number greater than zero... wouldn't the series diverge? My Professor said it congerves...but the p-series tests says it diverges when you compare it to 1/n...that is what is confusing me

Answer 28

No, because we're not taking the limit of just one sequence, we're taking the limit of the ratio of two sequences... so that if the limit of the ratio is some positive number, then their convergences are tied to each other (ie, either both converge or both diverge) the proof is a bit extensive...

Answer 29

ok so if it's greater than zero, it converges? and if it's less than zero it would diverge?

Answer 30

A requirement for the limit comparison test is that both series are positive... so it's kinda impossible for the limit to be negative :3

Answer 31

ohhhh ok....so when would a series diverge?

Answer 32

And no, if the limit is greater than zero, then either BOTH series converge or BOTH diverge...

Answer 33

Here's how it works... you have a series... you suspect it's convergent (divergent). So you compare it to a known convergent (divergent) series, and use the limit comparison test...
If the limit is a positive number, then your original series mus have been convergent (divergent), since a positive limit implies both series to be either convergent or divergent.

Answer 34

ok i think i got it. Will you be here long? I think i'm going to need help in other tests. I'm reviewing for my test this wednesday

Answer 35

Probably not... sorry... I'm sure there are others here that can help you out :)

Answer 36

i hope so. Thank you for your help!