Question

On finding the interval of convergence for a power series I need help testing an endpoint. x= -1 for (1/n2^n)*(x-1)^n which would be -2^n/n2^n ....what test would I use? Please show work.

Answer 1

\[\large \lim_{n \rightarrow +\infty}\frac{1}{n^{2n}}(-2)^n\]\[\large =\lim_{n \rightarrow +\infty}\frac{(-2)^n}{n^{2n}}\]\[\large =\lim_{n \rightarrow +\infty}(-1)^n\left(\frac{2}{n^{2}}\right)^n\]

Answer 2

because this is an alternating series, it is convergent if \[\large \lim_{n \rightarrow +\infty}\left( \frac{ 2 }{ n^2 } \right)^n=0\]divergent otherwise.
LET \[\large y=\lim_{n \rightarrow +\infty}\left( \frac{ 2 }{ n^2 } \right)^n\]\[\large \ln y=\lim_{n \rightarrow +\infty} n\ln\left( \frac{ 2 }{ n^2 } \right)\]

Answer 3

what is the power series?

Answer 4

Sorry the denominator is not n^(2n), it's n2^n

Answer 5

power series is \[\sum(1/n2^n)*(x-1)^n\]

Answer 6

LOL. so much for the effort. gtg. have work.

Answer 7

I'm sorry thanks though

Answer 8

\[\sum \frac{(x-1)^n}{n2^n}\]
like this?

Answer 9

yes

Answer 10

\[\large \lim_{n \rightarrow +\infty}\frac{ (-2)^n }{ n 2^n }=\lim_{n \rightarrow +\infty}(-1)^n\frac{ 2^n }{ n 2^n }=\lim_{n \rightarrow +\infty}(-1)^n\frac{ 1 }{ n }=\ln 2\] CONVERGENT!

Answer 11

but what test is that?

Answer 12

the steps i take to an interval of convergence is
\[\lim_{n\to~inf}\frac{f(n+1)}{f(n)}\]
\[\lim_{n\to~inf}\frac{(x-1)^{n+1}}{(n+1)2^{n+1}}~\frac{n2^n}{(x-1)^n}\]
\[\lim_{n\to~inf}\frac{n~(x-1)^{n+1-n}}{(n+1)2^{n+1-n}}\]
\[\lim_{n\to~inf}\frac{n~(x-1)}{(n+1)2}\]
factor out all the no-n parts
\[\left|\frac{x-1}{2}\right|~\lim_{n\to~inf}\frac{n}{n+1}\]
\[\left|\frac{x-1}{2}\right|*1\]

Radiance of convergence is then
\[R:\left|\frac{x-1}{2}\right|<1\]
\[R:|x-1|<2\]
\[R:|x|<3\]

giving me an interval of convergence of : -3 < x < 3

Answer 13

I had an interval of convergence as -1<x<3 which the teacher said is right, the only issue i had was testing the endpoint of x=-1 and what test to use to do that.

Answer 14

took it a little too far :)

R: |x−1| < 2

therefore, -1 < x < 3

Answer 15

aternating series test \[\large \sum_{n=1}^{+\infty}(-1)^n a_n\]is convergent if \[\large \lim_{n \rightarrow +\infty}a_n=0\]

Answer 16

if it was AST then your answer would be divergent because it didn't equal 0....

Answer 17

well, if you can conform the power series to:\[\ln(\frac{2}{3-x})\] when x=-1 is not a bad construction

and the sum when x=-1 in the power series is just 0+0+0+0+... = 0

so it doesnt look like they reach the same value at x=-1 to me; but thats prolly not kosher

Answer 18

\[\lim_{n \rightarrow +\infty}\frac{ 1 }{ n }=0\]

Answer 19

\[\lim_{n \rightarrow +\infty}(-1)^n \frac{ 1 }{ n }=\ln 2\]

Answer 20

right so it diverges because it equals ln 2 not 0....but earlier you said that it equaled ln 2 and that converged

Answer 21

AST says it is convergent, and the series converges to ln(2)

Answer 22

no the AST says that the lim of Cn converges if it equals 0, not a number

Answer 23

@Autumn0909 you are confused. AST determines only the convergence of an alternating series. The AST tests the SEQUENCE by \[\large \lim_{n \rightarrow +\infty}c_n=0\] and concludes the convergence of the SERIES. It cannot determine the limiting sum of the series.