Question

I'm needing help (refresher) on power series and series in general. Calc II or Differential Equations, either way.

First problem: Find all values of x for which this series converges:
∑(4^n)(x^n)(n+1)/(n+9); n0=1

Answer 1

I'd rather if you said how to do it, rather than the answer.

Answer 2

\[\large \sum_{n=0}^{\infty}\frac{(4^n)(n+1)(x^n)}{(n+9)}\]

is that correct?

Answer 3

yeah, exactly right.

Answer 4

sorry, i should have posted the pic

Answer 5

Unfortunately I don't have my calc textbook, or it wouldn't be a problem. :(

Answer 6

The ratio test would take care of this i believe. Let:\[a_n=\frac{4^n\cdot x^n(n+1)}{n+9}\] and you want to figure out when:\[\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|<1\]

Answer 7

ok, so it would be for values of x that make that true.

Answer 8

right.

Answer 9

I hate series sooo much >.<

Answer 10

Basically we end up with the power series converging if:\[|4x|<1 \Longrightarrow |x|<\frac{1}{4}\]

Answer 11

lol! I actually got that far before, just didn't know where to go from there.

Answer 12

my differential equations book doesn't cover series very well

Answer 13

and yeah, I find series a pain, too.

Answer 14

you gotta go back to a Calculus book rather than a Diff Eqn book. Thats what i had to do lol.

Answer 15

let's see if The Calculus Lifesaver says enough… :D

Answer 16

hm… "it" (online homework software) doesn't like [-1/4,1/4] as the interval of convergence

Answer 17

try ( , ) instead of brackets. The brackets give the impression that its ok if x = 1/4, but thats not the case.

Answer 18

it doesn't take that either. (current answer is (-.25,.25)

Answer 19

i cant figure out why it wouldnt take that...

Answer 20

by the nth term test, I get (4^n)*(x^n)

Answer 21

me neither. Might this be a case of needing to apply multiple tests?