If a series is divergent... does that mean that the series approaches infinity?

Divergent just means that the series does not approach a finite number.. which ought to mean that it either approaches Positive Infinity or Negative Infinity.

Correct?


Note: Dont worry Sine and Cosine functions. As my function is 4(4/3)^x

diverges since 4/3 > 1

/edit: to answer your question, convergent means the sum of the sequence approaches a fixed number. divergent just means it's not convergent. if a series approaches infinity, then it is divergent. but i'm not convinced the converse is true.

This post was edited by carteblanche on Dec 11 2014 08:40pm

Divergent means that the series does not approach a limit. Therefore the series does not have to approach +infinity or -infinity to be divergent

ie. (-1)^n is divergent, however it does not approach infinity

4(4/3)^x is divergent by the geometric test

http://www.stewartcalculus.com/data/BRIEF%20APPLIED%20CALCULUS/upfiles/challenge/ess_cp_08_stu.pdf

Im trying to help explain how 5(b) was found.

link does not work

That is strange that your link is different from mine. Mine has % signs, yours does not.

Link works if you copy and paste it into address bar.

It is the first hit that comes up on Google, if this method brings up a Google Search

This post was edited by CamelFinger on Dec 11 2014 09:18pm

Geometric Test cant be used because 4/3 is larger than 1 (which is a req of using this test)

explain the bolded. since it's larger than 1, it diverges, which is exactly what the geometric test says.

Sorry I misread my cheat sheet, you are right.

Since the common ratio r is 4/3 (which is greater than 1), the series Diverges.

Is it enuf to say that "the series approaches infinity because the series diverges according to the Geometric Test"? Or can a series be divergent, yet not approach infinity?

This post was edited by CamelFinger on Dec 11 2014 09:54pm

2wo1ne and i both said it already. if it approaches infinity, it is divergent; the converse is not true.

and you got your statement backwards. i assume you're trying to prove it's divergent? so you say it's divergent because the series approaches infinity, not the other way around. it's very obviously approaching infinity because each term is getting larger by a common ratio.

This post was edited by carteblanche on Dec 11 2014 09:58pm