Find a power series for the function

[arl146]

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

 

[Dick]

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

Why haven't you at least tried to put the series in a simpler form? Like by using 3^(2n)=9^n??

 

[arl146]

uhhh i don't know .. what does that help?

 

[Dick]

uhhh i don't know .. what does that help?

Simplifying a complicated expression usually helps you answer questions about it.

 

[arl146]

well i don't know, seeing it at 3^(2n) doesn't really make anything click in my head anymore than 9^n does

 

[Dick]

well i don't know, seeing it at 3^(2n) doesn't really make anything click in my head anymore than 9^n does

There's a 3^(2n) in the numerator and there's a 9^n in the denominator. If you cancel them then you don't have to look at either one. I'm not sure what this debate about "why should I simplify" is accomplishing.

 

[arl146]

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?

 

[Dick]

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?

Use algebra to figure out how 3^(2n+1) is related to 3^(2n). Are you doing this on your phone again or something? Is that what's going on with this short attention span thing? Get a piece of paper and work it out.

 

[arl146]

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

 

[Dick]

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

1 is in the exponent of something. It's not just 1. This is getting pretty painful.

 

[arl146]

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)) ?

 

[Dick]

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)) ?

They DO cancel. Why didn't you cancel them before? So you would get 1/(2n+1) if you weren't forgetting something else.

 

[arl146]

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?!

 

[Dick]

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?!

Not with the simplification, you just left a factor behind several posts ago. Like usual.

 

[arl146]

other than the (-1)^n i don't see anything

 

[Dick]

other than the (-1)^n i don't see anything

That's the one. So does it converge or diverge at the endpoints?

 

[arl146]

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: b_n+1 <= b_n

b_n+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

 

[Dick]

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: b_n+1 <= b_n

b_n+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

Ok, sorry about the 'as usual'. But after 50+ posts on this I'm getting tired. Yes, it's convergent on both endpoints by the alternating series test. I could quibble about you leaving the (-1)^n in there when you are checking whether the absolute value of the terms are decreasing, but I'm going to let that one go.

 

[arl146]

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.

 

[Dick]

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.

You are certainly entitled to your opinion. You are welcome.