Question

Can anybody solve this limit of sequence problem??

Answer 1

this sequence alternates

are you trying to look a sequence or a series maybe to determine it's convergence or non-convergence ?

Answer 2

and what i mean by this sequence alternates like is that every other number is positive

Answer 3

em...I just want to figure out the limit of this

Answer 4

do you understand what I'm saying or what it means about this limit that I'm telling you that every other term is positive?

Answer 5

it wouldn't it matter maybe so much if that thing next to the thing that makes it alternate went to 0 but it does not

Answer 6

Sorry I still don't get what you mean....Can you draw something?

Answer 7

\[n=1 ;(-1)^{1-1} \cdot \frac{1+1}{1+2}=1 \cdot \frac{2}{3} \\ n=2; (-1)^{2-1} \frac{2+1}{2+2}=-1 \cdot \frac{3}{4} \\ n=3; (-1)^{3-1} \frac{3+1}{3+2}=1 \cdot \frac{4}{5} \\ n=4; (-1)^{4-1} \frac{4+1}{4+2}=-1 \cdot \frac{5}{6} \\ \cdots \\ n=100; (-1)^{100-1}\frac{100+1}{100+2}=-1 \cdot \frac{101}{102} \\ n=101; (-1)^{101-1}\frac{101+1}{101+2}=1 \cdot \frac{102}{103}\]
and so on
can you make any conjectures about this sequence now?

Answer 8

so they are approaching 0???

Answer 9

no

Answer 10

102/103 is closer to 1 not 0

Answer 11

but the sequence doesn't converge to 1 either because as I said the sequence is alternating

Answer 12

actually the denominator is n+1 and numerator is n+2 , so would it be closer to 0?

Answer 13

no

Answer 14

that is like saying 103/102 is closer to 0

Answer 15

102/102 is 1
so if you have 103/102 that is going to be a little greater than 1

Answer 16

yes

Answer 17

it didn't matter if you had (n+2)/(n+1) or if you had (n+1)/(n+2)
either of those second parts it is getting closer to 1

what kills everything is the alternating part

Answer 18

oh I get it :)

Answer 19

if you had
\[\lim_{n \rightarrow \infty}(-1)^{n-1} \frac{1}{n}\]
this would go to zero
do you know why?

Answer 20

because 1/n is 0 so the whole thing is 0?

Answer 21

well yes 1/n goes to zero so the alternating part doesn't matter
because 0*whatever is 0

Answer 22

well let me rephrase that whatever when whatever is defined :p

Answer 23

0*infinity is an indeterminate form which will totally see later
you cannot just say this form is zero :p

Answer 24

ok I get this part, but I am just wondering what could (-1)^(n-1) be?

Answer 25

that is the thing that cases it to alternate

Answer 26

Let g(n)=(-1)^(n-1)
then for example g(1)=1
and g(2)=-1
and g(3)=1
and g(4)=-1
and g(5)=1
...
basically g(2n)=-1 where n is an integer
and g(2n+1)=1 where n is an integer

by the way 2n represents all the even integers (where n is integer)
and 2n+1 represents all the odd integers (where n is integer)

Answer 27

so if i want to determine \[\lim_{n \rightarrow \infty} (-1)^{n-1}\]
what would that be?
I got 2n and 2n+1 part..:)

Answer 28

well again do you see these numbers getting close to anything
1,-1,1,-1,1,-1,...
like you would have the only choices 1 or -1
but it cannot be either because your sequence here is not getting closer to any ONE number

Answer 29

just like the question before you would say the limit does not exist
or the sequence does not converge

Answer 30

so because it actually "has two limits", then it doesn't have limit. Can I say that?

Answer 31

yes I think I get that
(though it can't really have two limits)
hmmm.... you know what I think there is a sub-sequence theorem
@rational correct me if I'm wrong (and if you know)
if you look at infinite subseqeunces (and say you found only two ) in your seqeunce and they converge to two difference numbers then sequence for which you found those infinite subsequences diverges

so for example we had
\[g(n)=\left\{ 1,-1,1,-1,1,-1,... \right\} \\h(n)= g(2n)=\left\{ -1,-1,-1,-1,... \right\} \\ f(n)=g(2n+1)=\left\{ 1,1,1,1,1,1,.... \right\} \\ \lim_{n \rightarrow \infty}h(n)=-1 \\ \text{ and } \lim_{n \rightarrow \infty}f(n)=1 \\ \text{ so } \\ \lim_{n \rightarrow \infty} g(n) \text{ does not converge }\]

Answer 32

http://www.maths.manchester.ac.uk/~mdc/old/153/notes3.pdf ha found it
corollary 3.5

Answer 33

on page 8

Answer 34

So for my question, because the sequence is divergent, it doesn't have limit.

Am I correct?

Answer 35

oh yes totally

well there are difference types of divergence

you can say the limit does not exist or goes to infinity or neg infinity depending on what the numbers are doing

in your case the limit would not exist (these numbers do not get super large or super negative large )

Answer 36

Oh I eventually got it!
Thank you so much for the patience...

Answer 37

np

Answer 38

Hey also I shouldn't have put { } around my sequence of numbers. That is the only thing I want to change or omit about this thread.