Mathematics OpenStudy (anonymous):

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

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

at first glance i would say this converges for all real numbers OpenStudy (anonymous):

can u help me prove it by teh ratio test....thats teh wrk i need to show OpenStudy (anonymous):

maybe i need to think before i speak, but you have an alternating sum OpenStudy (anonymous):

is it becuase its -1 to a power? OpenStudy (anonymous):

hold on i need to think OpenStudy (anonymous):

ok no problem OpenStudy (anonymous):

scratch that entirely OpenStudy (anonymous):

$\sum(-\frac{1}{3})^k\frac{x^k}{k+1}$ is what we need to consider OpenStudy (anonymous):

ok i see that OpenStudy (anonymous):

i think ratio test will show that this converges if $|x|<3$ OpenStudy (anonymous):

how did u get that....dont u need to show a sub n and sub n+1 and tehn do limits OpenStudy (anonymous):

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 OpenStudy (anonymous):

yeah ratio test does it OpenStudy (anonymous):

$\frac{a_{k+1}}{a_k}=-\frac{3^k(k+1)}{3^{k+1}(k+2)}$ $=\frac{k-1}{3(k+2)}$ take the limit and get $\frac{1}{3}$ OpenStudy (anonymous):

oh i may be off by a minus sign, but no matter OpenStudy (anonymous):

does this mean that $-1/3 < x < 1/3$ OpenStudy (anonymous):

cuz i know my teach finds the Raduis of convergence then test each to see if its included OpenStudy (anonymous):

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 OpenStudy (anonymous):

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 OpenStudy (anonymous):

so for example if the limit was 2 we would have to take $|x|<\frac{1}{2}$ OpenStudy (anonymous):

if the limit is zero, it converges for all x OpenStudy (anonymous):

oh ok so if limit is 1/3 it would be absolute value of x less than 3 OpenStudy (anonymous):

and if the limit does not exist, it only converges at x =0 OpenStudy (anonymous):

yes you are right OpenStudy (anonymous):

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 OpenStudy (anonymous):

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 OpenStudy (anonymous):

yes, and you may be required to test at the endpoints of the interval OpenStudy (anonymous):

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 OpenStudy (anonymous):

yeah so wen u subsitite -3 into the limit does it converge or diverge? OpenStudy (anonymous):

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

may u plz show me how OpenStudy (anonymous):

using limits OpenStudy (anonymous):

ok first we can do it with x = 3 and show that it works OpenStudy (anonymous):

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 OpenStudy (anonymous):

problem is if you replace x by -3 it does not alternate OpenStudy (anonymous):

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 OpenStudy (anonymous):

and that one clearly does not converge OpenStudy (anonymous):

so if it doesnt converge, it diverges right? OpenStudy (anonymous):

yes so the actual inteval of convergence is $-3<x\leq 3$ OpenStudy (anonymous):

-3 does not converge, 3 does. k? OpenStudy (anonymous):

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

yw

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