Question

need help finding the interval of convergence for a power series.

Answer 1

I know the radius of convergence is R=1

Answer 2

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

Answer 3

the ratio test gave me \[\lim_{n \rightarrow \infty}\left| \frac{ a _{n+1} }{ a _{n} } \right|=1\] and this is an inconclusive statement for the ratio test. From this point I honestly don't know what to do to figure out the convergence or divergence of the endpoints

Answer 4

The limit is not actually 1:
\[\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\left|\frac{x^{n+2}}{(n+1)(n+2)}\cdot\frac{n(n+1)}{x^{n+1}}\right|=x\lim_{n\to\infty}\left|\frac{n}{n+2}\right|=x\]
Note that you're given a geometric series, which means it converges for \(|x|<1\), i.e. \(-1<x<1\).

Since the length of this interval is 2, the radius is 1, like you said earlier.

Answer 5

Sorry, should be \(|x|\) as the limit.

Answer 6

I am kind of new at this part. I thought once you knew the radius of convergence you plugged x in to find the endpoint convergence. And I got \[\left| x \right|\]as the limit as well, but when I plugged in the one and negative one for the endpoints because the radius is centered at x=0 I got the absolute value was equal to one. Is that the wrong approach?

Answer 7

@SithsAndGiggles ?

Answer 8

@zepdrix ?

Answer 9

@dumbcow ?

Answer 10

do not use ratio test to determine convergence/divergence at x=-1 or x = 1. Ratio Test is inconclusive.

Answer 11

when x=-1, the series is \[\sum \frac{(-1)^{n+1}}{n(n+1)}\] is convergent by the alternating series test.

Answer 12

what test can you use to determine convergence/divergence at x = 1?

Answer 13

honestly I don't know I didn't even think of the alternating series test for the negative one. would it be the direct comparison test?

Answer 14

direct comparison test will work for the case x = 1.

Answer 15

and by the alternating series test I would only have to evaluate \[\lim_{n \rightarrow \infty}\frac{ 1 }{ n(n+1) }\] ?

Answer 16

direct comparison test means you have to compare your series-problem to some standard series that is known to be convergent or divergent.

Answer 17

in this case, \[\frac{1}{n(n+1)}=\frac{1}{n^2+n}<\frac{1}{n^2}\] the standard series for comparison is \[\sum \frac{1}{n^2}\] which is known to be convergent

Answer 18

because it is a p series where p>1

Answer 19

that makes sense. In the alternating series, when evaluating the limit of the a sub nth term I just exclude the (-1)^(n+1) right?

Answer 20

yup, that's correct. now use the rules described in the comparison test, and conclude that the series-problem is convergent/divergent.

Answer 21

I am rusty on all the infinite series tests sorry...

Answer 22

you're also correct in the alternating series test, the (-1)^(n+1) does not appear in the limit evaluation process.

Answer 23

so with an alternating series I just have to prove that the limit equals zero and that the a sub n+1 st term is less than the a sub nth term?

Answer 24

if that is correct just one other quick question..... why do we have to prove the a sub n+1 st term is less than the a sub nth? If I understand why it helps me remember?

Answer 25

consider the sequence of positive terms an , \[a_{n+1} < a_n\] then it follows that
\[\frac{a_{n+1}}{a_n}<1\] and this is the criterion for the ratio test.

Answer 26

thanks