Question on limits of alt. signs

[nuclearrape66]

If I'm taking the limit of a sequence of numbers that is {An} and An alternats in signs.
am i suppose to take the absolute value to determine the limit. and if the absolute value of An diverges then the lmit diverges?

or how am i suupose to take the limit of alternating signs?

 

[sutupidmath]

well, the sequence probbably alternate its sign as you plug in an even or an odd number. So what u need to do is try first to redefine n as only an even number that is of the form n=2k, where k is an integer, and the other time define n to be only odd, that is n=2k-1. an look what happenes when you take the limit? If the limit for n even and for n odd is the same than the series converges but if the limit as n is odd and as n is even are not the same, then the sequence does not have a limit, so it diverges.
Do you understand what you need to do?

 

[nuclearrape66]

so youre saying to plug in 2n for n and get its limit...


then plug in 2n + 1 for n and find its limit...

and if they both go towards the same thing then it converges to that number?

 

[sutupidmath]

well, you can let n=2k, and then as n-->infinity, also k-->infinity, the same with
n=2k-1, as n-->infinity, also k--->infinity? Then if the both limits go to the same number the overall limit of that sequence will be that particular number, in contrary the limit does not exist.

 

[nuclearrape66]

so for (-1)^n(1/n) i would do:

lim (-1)^2n(1/2n) = 0

and

lim (-1)^2n+1(1/2n + 1 ) = 0

and therefore it's 0.

but what about (-1)^n(n/n+2)

 

[sutupidmath]

well the last one does not have a limit, since when n:=2n, and when n:=2n+1, you will get
1, and -1 respectively.

 

[nuclearrape66]

oh i see, so so then if you do the absolute value thing, where you take the limit of any alternating sequeence and if that diverges then the alt sequence diverges if it converges then the sequence converges.

does that work all the time for these?