Find a power series for the function

[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.