Question

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)]

Answer 1

try ratio test, think it is
\[|x|<1\]

Answer 2

but what will be my a sub n+1

Answer 3

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

Answer 4

you have
\[sum\frac{(-1)^{k+1}x^k}{k}\]

Answer 5

so ratio is
\[\frac{k}{k+1}\] limit is 1

Answer 6

therefore radius of convergence is
\[|x|<1\] want to try at the endpoints?

Answer 7

yes

Answer 8

one will work and the other will not, depending on whether you get an alternating sum or not

Answer 9

so at which one is it divergent

Answer 10

satellite could you help me with that one probelm again i had a question :0

Answer 11

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

Answer 12

but at minus one the series will not alternate

Answer 13

um b4: \[\frac{k}{k+1}\] limit is 1
how did u get that

Answer 14

limit as k goes to infinity? it is clearly 1 right?

Answer 15

hold on let me think before i type

Answer 16

ok it is right,what i wrote

Answer 17

if you replace x by -1 it will not converge

Answer 18

why is that and which series test proves the divergence

Answer 19

\[\sum\frac{(-1)^{k+1}(-1)^k}{k}\]
\[\sum\frac{(-1)^{2k+1}}{k}\]
\[\sum\frac{-1}{k}\] does not converge

\[2k+1\] is odd and therefore
\[(-1)^{2k+1}=-1\]

Answer 20

\[\sum\frac{-1}{k}=-\sum\frac{1}{k}\]diverges because it is the harmonic series

Answer 21

so the negative can just be taken out?

Answer 22

sure why not? it is a constant

Answer 23

ohhh ok i see. Thanks once again!!!

Answer 24

yw