Question

Need help understanding the concept.
Does \(\dfrac{(-1)^n}{n}\) converge or diverge?

Answer 1

Let \(X_n= \dfrac{(-1)^n}{n}\)

Answer 2

\(X_n= (-1)^n *\dfrac{1}{n}\) so that \(lim X_n =lim (-1)^n *lim 1/n = 0 \)
Which shows it converges. Am I right?

Answer 3

One more thing, if Xn diverges , is it possible that it is bounded and have lim?
for example \(X_n = (-1)^n \) it is diverge, but it is bounded by -1, and 1,

Answer 4

it converges for sure, but not the limit test is not useful. you have used alternating series test, right ?

Answer 5

A series that doesn't have a limit it converges, so yeah it's possible for it to be bounded.

Answer 6

Diverges*

Answer 7

I don't use test to know whether it converges or not, because it is trivial that lim =0 :)

Answer 8

sequence converges does not imply the series converges

Answer 9

a sequence converges --> it has a limit, right?

Answer 10

yes oh you're talkign about sequence.. then fine :)

Answer 11

looks i have misread the question as series hmm

Answer 12

What do you mean by "opposite"?

Answer 13

I though you talked about series

Answer 14

I meant the other way arround

Answer 15

@ganeshie8 I confuse between conve/diverge of sequence and series, too. But for now, it is a sequence, solve it first. I will ask for series later

Answer 16

you have already solved it right, whats there to solve further ?

Answer 17

@math&ing001 a sequence converges--> has limit
a sequence diverges --> no bound --> no limit, right?

Answer 18

Yeah, that's right.

Answer 19

how about \(X_n= (-1)^n\), it diverges but bounded by -1,1

Answer 20

and supremum is 1, infimum is -1.

Answer 21

Yes that's was I was talking about earlier, it may be bounded but still diverges because it has no limit.

Answer 22

You see, it confuses me, earlier: diverges --> no bound--> no limit
this case: diverges --> but bounded. ha??

Answer 23

I didn't pay attention to your post, this should be removed :
a sequence converges--> has limit
a sequence diverges -->(( no bound -->)) no limit

Answer 24

By the way, I think you're talking about series here.
This is the sequence (-1)^1 , (-1)^2/2 , (-1)^3/3 , .....
This is the series (-1)^1 + (-1)^2/2 + (-1)^3/3 + .....

Answer 25

I know the definition of series, it just partial sum of sequence. However, all concept about series come from sequence. So that I want to make it clear before moving to the series.
Obviously, the sequence (-1)^n has 2 converged subsequences (1)^n and (-1)^n where n even and odd respectively. But itself (-1)^n is diverges. And I confused about its bounds

Answer 26

Or: bounded sequence doesn't mean the sequence convergent. ?
And it confuses me also. Ha!!

Answer 27

Yeah I understand the confusion, just remember this :
-1<(-1)^n<1
=> \[\sum_{}^{}(-1) \le \sum_{}^{}(-1)^{n} \le \sum_{}^{} 1\] and technically both those series diverge so it doesn't mean anything.