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

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

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

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

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

is it becuase its -1 to a power?

hold on i need to think

ok no problem

scratch that entirely

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

ok i see that

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

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

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

yeah ratio test does it

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

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

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

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

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

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

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

if the limit is zero, it converges for all x

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

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

yes you are right

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

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

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

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

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

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

may u plz show me how

using limits

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

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

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

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

and that one clearly does not converge

so if it doesnt converge, it diverges right?

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

-3 does not converge, 3 does. k?

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

yw

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