arl146 said:Yea there's only 2 examples and they don't have any x's involved like mine. The book really doesn't show good examples in my opinion for reasons like that. The one examples is "test the series summation ((-1)^n * n^3)/(3^n) for absolute convergence." and all they have to do is do the ratio test, in which they get the limit equal to 1/3 which is obviously less than 1. The other examples is the same thing!
Anyways, the a sub n+1 part would equal x^(2(n+1)+1) on top so just x^(2n+3) and 3*(2(n+1)+1)*(9^n+1) on bottom. And that is multiplied times the a sub n part which is 3*(2n+1)*(9n) on top and x^(2n+1) on bottom. Sorry I can't throw this stuff all together in a better way, still on my phone. I simplified all that to ((x^2)*(2n+1))/(9*(2n+3)) and that's how I got my first part I gave you
arl146 said:Negative 3 and positive 3. Is this where I have to then check the end points which is just +3 and -3 like I said?
arl146 said:Ok and so for that I just have to plug in the endpoints into the series we found and I have to use the alternating series test for the convergence test of the endpoints right?
arl146 said:ok so for x=3,we have summation (-1)^n * ((3^(2n+1))/(3*(2n+1)*(9^n)))
and for x=-3 we have summation (-1)^n * (((-3)^(2n+1))/(3*(2n+1)*(9^n)))
dont i have to use alternating series for this series though?
for the alternating series test the series has to satisfy 2 things:
(i.) the series with n+1 in replacement of all n's has to be <= the series when n just equals n
(ii.) limit of the series has to equal 0.
could you help me with completing this limit? i usually tend to go towards L'Hopitals but i know in this case that wouldn't be a good route to take so how do i do the limit? i haven't done limits in quite some time. i know that i can take the limit of the top and bottom 'individually' so starting with the top, the x^(2n+1) part, am i able to just plug in n=infinity for the n? and if that's the case .. for the bottom, do i break that up into 2parts, the (2n+1) and the 9^n ? since i can just 'pull' the 1/3 out in front. and once i separate the bottom can i plug in the n=infinity into that too?
as for part (i.) ... i also don't exactly know how to tell if one is less than or greater than the other ... someone else was helping me with a problem like this and i just didnt understand
arl146 said:uhhh i don't know .. what does that help?
arl146 said:well i don't know, seeing it at 3^(2n) doesn't really make anything click in my head anymore than 9^n does
arl146 said:ohhhh yea i forgot that i plugged in the 3 for the n on top that's why i was confused. but its a 3^(2n+1) so how do i cancel them.. I am just left with 1 on top and bottom then, right? so then I am left with it being 1/(3*(2n+1)) well really i have (1/3)*lim 1/(2n+1) right?
arl146 said:yea i know, i just subtract the exponents. 2n+1-2n youre just left with 1. which is what i said. i mean unless that's wrong but that's what I've been doing and it worked for previous stuff
arl146 said:i know i know sorrrrrry. i was thinking that the 3's cancel but they dont. so I am left with 3/(3*(2n+1)) ?
arl146 said:yea i saw that but i was talking about something that is irrelevant at this point.
i see the 1/(2n+1) .. and for my endpoints one is + and the other -
im forgetting something else with simplifying?!
arl146 said:other than the (-1)^n i don't see anything
arl146 said:ok, well the 'as usual' isn't even necessary, we were working on simplifying i wasnt concerned with that at the moment i didnt forget it, geez.
well, part 2 of the alternating series test is satisfied, the limit equals 0.
for part 1, it has to satisfy this: bn+1 <= bn
bn+1 = (-1)^n * \frac{1}{2n+3}
and so (-1)^n * \frac{1}{2n+3} is < (-1)^n * \frac{1}{2n+1}
so that is satisfied too .. so convergent on both endpoints
arl146 said:i think i am done with the problem then. the radius of convergence is 3?
im just saying, be more patient. people come here for help, not to feel stupid by the 'more intelligent' helpers. i appreciate your help but i definitely don't appreciate your attitude, and i shouldn't have to put up with it just because I am getting free help. thanks.