can I do this w/ my proof?

[Semidevil]

so given ((-1)^n)*n/ (n + 1), for all n in Natural numbers.

so I need to prove that it converge or diverge. so i break this up.

well...we know that ((-1)^n) diverges. I multiply top and bottom by 1/n, and the new fraction becomes ((-1)^n)/(1 + 1/n). I'm confused. So I know the top will diverge, but the bottom will converge to 1.

So does this converge or diverge? is there a rule about a fraction of diverge/converge?

oh yea, this is not calculus, only analysis.

 

[notfred]

nm

 

[Semidevil]

Originally posted by: notfred
nm

interesting....

so can I make my equation to become (-1)^n/1 and say it diverges?


but wont this mean that i'm not takig the limit of both at the same time?? is tha allowed?

 

[mchammer187]

it doesnt do either

there is an upper and lower bound though

it looks like a sine wave IIRC

 

[mygumballs]

is this an infinite series of just a sequence? are we summing this thing? because if it's not a series and just a sequence, then the limit as n tends to infinity is some function that alterates between 1 and negative one.

 

[Semidevil]

dont get it....

so I know that the (-1)^n will diverge for sure...between -1 to 1, or -n to n. but....what do I do with the bottom term?? does that change anything?

was my method correct?

 

[Semidevil]

Originally posted by: mygumballs
is this an infinite series of just a sequence? are we summing this thing? because if it's not a series and just a sequence, then the limit as n tends to infinity is some function that alterates between 1 and negative one.

this is just a sequence....


so how do you actually prove that it doesn't converge?

 

[TuxDave]

I take it back.

 

[mygumballs]

Originally posted by: Semidevil

Originally posted by: mygumballs
is this an infinite series of just a sequence? are we summing this thing? because if it's not a series and just a sequence, then the limit as n tends to infinity is some function that alterates between 1 and negative one.

this is just a sequence....


so how do you actually prove that it doesn't converge?

like you said, you break it up. (-1)^n does not diverge, because, as mchammer pointed out, it is bounded above and below by 1 and -1 respectively. any bounded sequence must converge. and the other part is simply taking the limit of 1/(1 + (1/n)) = 1 which clearly doesn't diverge

 

[TuxDave]

Originally posted by: mygumballs

Originally posted by: Semidevil

Originally posted by: mygumballs
is this an infinite series of just a sequence? are we summing this thing? because if it's not a series and just a sequence, then the limit as n tends to infinity is some function that alterates between 1 and negative one.

this is just a sequence....


so how do you actually prove that it doesn't converge?

like you said, you break it up. (-1)^n does not diverge, because, as mchammer pointed out, it is bounded above and below by 1 and -1 respectively. any bounded sequence must converge. and the other part is simply taking the limit of 1/(1 + (1/n)) = 1 which clearly doesn't diverge

Wait... that's the definition of converge (to be bounded)? Heh.. I always thought it meant that it'll limit to some constant.

 

[Semidevil]

yea, I mean, if (-1)^n doesn't diverge, then what does it converge to??

 

[mygumballs]

Originally posted by: TuxDave

Originally posted by: mygumballs

Originally posted by: Semidevil

Originally posted by: mygumballs
is this an infinite series of just a sequence? are we summing this thing? because if it's not a series and just a sequence, then the limit as n tends to infinity is some function that alterates between 1 and negative one.

this is just a sequence....


so how do you actually prove that it doesn't converge?

like you said, you break it up. (-1)^n does not diverge, because, as mchammer pointed out, it is bounded above and below by 1 and -1 respectively. any bounded sequence must converge. and the other part is simply taking the limit of 1/(1 + (1/n)) = 1 which clearly doesn't diverge

Wait... that's the definition of converge (to be bounded)? Heh.. I always thought it meant that it'll limit to some constant.

well, it's not the definition of convergence, it's just a theroem that results from it.
the definition of covergence is for some large N, and for any e > 0, n >= N implies the distance between the sequence and p are less than e, where p is the point the sequence converges to

 

[mchammer187]

Originally posted by: Semidevil
yea, I mean, if (-1)^n doesn't diverge, then what does it converge to??

it doesnt

that is like saying lim (sin(x)) : x--> infinity

it does not converge or diverge

it stays between two numbers -1 and 1

just like your first limit

 

[Semidevil]

ok crap, i'm getting confused. ok, so (-1)^n doesn't converge or diverge....


so back to my original problem, how do I prove that it whole thing converge or diverge?

 

[TuxDave]

Originally posted by: mchammer187

Originally posted by: Semidevil
yea, I mean, if (-1)^n doesn't diverge, then what does it converge to??

it doesnt

that is like saying lim (sin(x)) : x--> infinity

it does not converge or diverge

it stays between two numbers -1 and 1

just like your first limit

http://mathworld.wolfram.com/ConvergentSequence.html

Uses the definition that I know. You're either convergant or divergant. There is no 'other'.

'If does not converge, it is said to diverge.'

 

[DrPizza]

why not show that the n/(n+1) part converges to 1...
so, you end up with -1^n, which neither converges nor diverges; it oscillates between -1 and 1

 

[mygumballs]

Originally posted by: Semidevil
ok crap, i'm getting confused. ok, so (-1)^n doesn't converge or diverge....


so back to my original problem, how do I prove that it whole thing converge or diverge?

another way to tackle the problem is to split the sequence into 2 subsequences of (-1)^n for all odd n or all even n. both of these subsequences obviously converge to -1 and 1 respectively. there should be some theorem that says if all of the subsequences that make up a sequence coverge, then the sequence converges.

the sequence doesn't necessarily need to converge to one value. it ca converge to a bounded function