Question

Trying to find the interval of convergence for the sum from n=1 to INF of [(x-3)^n]/[n(-5)^n]. I've made a few attempts (see attached image) and keep getting the same (wrong) result. Thanks. :)

Answer 1

did you check your endpoints?

Answer 2

Yeah, that's what I was writing in the part that says "for x=-2" and "for x=8". I'm pretty sure I'm not doing it *right*. :)

Answer 3

what happens when x=8

Answer 4

When x=8, I get\[\lim_{n \rightarrow \infty} \frac{ (5)^{n}}{ n(-5)^{n} }\]

Answer 5

what happened to the sum?

Answer 6

I was checking for convergence.

Answer 7

\[\large\sum_{n=1}^{\infty}\frac{5^n}{n(-5)^n}\]

Answer 8

\[=\large\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\]

Answer 9

Ok, I had it that far, although in a limit. Ergh... I guess showing that it *doesn't* diverge by limit going to 0 isn't enough to say that the sum on x=8 converges?

Answer 10

no...you need to use the alternating series test

Answer 11

Ah. Ok, give me a sec to dig that up (lot to memorize in an accelerated summer class).

Answer 12

So if {a_n} is decreasing, and going to zero in the limit, the series converges. Is that right?

Answer 13

if {a_n} is decreasing, and going to zero in the limit,

then

\[\sum_{n=1}^{\infty}(-1)^na_n\] converges

Answer 14

So we'd say that the right side of the interval of convergence includes 8. Ok, I need to try that again for -2, then.

Answer 15

correct

Answer 16

So the left side comes from the harmonic sum? \[\sum_{1}^{\infty} \frac{1}{n}\]

Answer 17

correct

Answer 18

and what does the harmonic do

Answer 19

It diverges?

Answer 20

correct
so what is your interval of convergence?

Answer 21

(-2,8]?

Answer 22

you got it

Answer 23

Let me type that in again...Webwork is kind of testy.

Answer 24

That was it, thanks! I can tell I need a lot more practice. Thanks so much for your help.

Answer 25

np