What are all the values of x for which the series converges x - (x^2/2) + (x^3/3) - (x^4/4)+...[(-1)^n+1* (x^n/n)]
try ratio test, think it is \[|x|<1\]
but what will be my a sub n+1
i was pretty sure it was from -1 to 1 but im not sure how 2 prove it or which of the 2 are included in the interval of convergence
you have \[sum\frac{(-1)^{k+1}x^k}{k}\]
so ratio is \[\frac{k}{k+1}\] limit is 1
therefore radius of convergence is \[|x|<1\] want to try at the endpoints?
yes
one will work and the other will not, depending on whether you get an alternating sum or not
so at which one is it divergent
satellite could you help me with that one probelm again i had a question :0
it will converge is x = 1 because you just get \[\sum\frac{(-1)^{k+1}}{k}\] an alternating sequence where the terms go to zero, so it will converge
but at minus one the series will not alternate
um b4: \[\frac{k}{k+1}\] limit is 1 how did u get that
limit as k goes to infinity? it is clearly 1 right?
hold on let me think before i type
ok it is right,what i wrote
if you replace x by -1 it will not converge
why is that and which series test proves the divergence
\[\sum\frac{(-1)^{k+1}(-1)^k}{k}\] \[\sum\frac{(-1)^{2k+1}}{k}\] \[\sum\frac{-1}{k}\] does not converge \[2k+1\] is odd and therefore \[(-1)^{2k+1}=-1\]
\[\sum\frac{-1}{k}=-\sum\frac{1}{k}\]diverges because it is the harmonic series
so the negative can just be taken out?
sure why not? it is a constant
ohhh ok i see. Thanks once again!!!
yw
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