Question

Determine the interval of convergence of the power series given: I will write the series w/ the equation box in a sec

Answer 1

\[\sum_{k=0}^{\infty} [(-1)^k x^k] / [ 3^k (k+1)]\]

Answer 2

at first glance i would say this converges for all real numbers

Answer 3

can u help me prove it by teh ratio test....thats teh wrk i need to show

Answer 4

maybe i need to think before i speak, but you have an alternating sum

Answer 5

is it becuase its -1 to a power?

Answer 6

hold on i need to think

Answer 7

ok no problem

Answer 8

scratch that entirely

Answer 9

\[\sum(-\frac{1}{3})^k\frac{x^k}{k+1}\] is what we need to consider

Answer 10

ok i see that

Answer 11

i think ratio test will show that this converges if
\[|x|<3\]

Answer 12

how did u get that....dont u need to show a sub n and sub n+1 and tehn do limits

Answer 13

let me write it with pencil before i type, but i think if you take the limit of the ratio you will get 1/3 which tells you x has to be less than 3

Answer 14

yeah ratio test does it

Answer 15

\[\frac{a_{k+1}}{a_k}=-\frac{3^k(k+1)}{3^{k+1}(k+2)}\]
\[=\frac{k-1}{3(k+2)}\] take the limit and get
\[\frac{1}{3}\]

Answer 16

oh i may be off by a minus sign, but no matter

Answer 17

does this mean that
\[-1/3 < x < 1/3\]

Answer 18

cuz i know my teach finds the Raduis of convergence then test each to see if its included

Answer 19

need absolute value anyway i think

no that means since we want this to be between -1 and 1 we can take x between -3 and 3

Answer 20

once we get the limit is 1/3 we are allowed to go up the the reciprocal, since we need to be between -1 and 1

Answer 21

so for example if the limit was 2 we would have to take
\[|x|<\frac{1}{2}\]

Answer 22

if the limit is zero, it converges for all x

Answer 23

oh ok so if limit is 1/3 it would be absolute value of x less than 3

Answer 24

and if the limit does not exist, it only converges at x =0

Answer 25

yes you are right

Answer 26

if you want to do a little experiment, replace the "x" here by various numbers and see what you get
http://www.wolframalpha.com/input/?i=sum+%28%28-1%29^k+x^k%29%2F%283^k%28k%2B1%29%29

Answer 27

so int of convergence is \[-3 <x <3\]
and we wud have to test each w/ teh limit to see if it was included in the interval of convergence

Answer 28

yes, and you may be required to test at the endpoints of the interval

Answer 29

because it may be
\[-3<x\leq 3\] or
\[-3<x<3\] or \[-3\leq x \leq 3\] etc
you don't know in advance

Answer 30

yeah so wen u subsitite -3 into the limit does it converge or diverge?

Answer 31

i just ran a little experiment and saw that 3 works and -3 does not
we could prove it though i am sure

Answer 32

may u plz show me how

Answer 33

using limits

Answer 34

ok first we can do it with x = 3 and show that it works

Answer 35

you get
\[\sum\frac{(-1)^k3^k}{3^k(k+1)}\]
\[\sum \frac{(-1)^k}{k+1}\] and this has to converge because it is an alternating series and the terms go to zero

Answer 36

problem is if you replace x by -3 it does not alternate

Answer 37

you get
\[\sum\frac{(-1)^k(-3)^k}{3^k(k+1)}\]
\[\sum\frac{(-1)^{2k}3^k}{3^k(k+1)}\]
\[\sum \frac{(-1)^{2k}}{k+1}\]
\[\sum\frac{1}{k+1}\] because
\[(-1)^{2k}=1\] always

Answer 38

and that one clearly does not converge

Answer 39

so if it doesnt converge, it diverges right?

Answer 40

yes
so the actual inteval of convergence is
\[-3<x\leq 3\]

Answer 41

-3 does not converge, 3 does.
k?

Answer 42

NICE!! THANKS SOO MUCH!!! It makes more sense wen i see it step by step instead of teh teacher zooming thru a problem

Answer 43

yw