Why does the series of (-1)^(n-1)x^n/n converge for x=1 but not -1

for n=0 to infinity of the series (-1)^(n-1)x^n/n
I need to find what x values the power series converges for.
I did the ration test for the absolute value and found the it absolutely converges from -1 to 1. Then I went to check the end points of the interval which are 1 and -1. I thought they both converged, but the answer says only x=1 converges. I don't understand why.

Wouldn't they be (-1)^(2n-1)/n and (-1)^(n-1)/n which both conditionally converge by alternating series test?

I had a similar problem with the series from n=1 to infinity of (-1)^(n-1)/(2n-1) I got that it converges from -1 to 1
but when I went to check endpoints this time both converged conditionally.
I don't understand.

Thanks you

OK set x= -1 then:

(-1)^(n-1) x^n /n = (-1)^(n-1) (-1)^n /n

Now by law of exponents:

(-1)^(n-1) (-1)^n = (-1)^(n-1 + n) = (-1)^(2n-1)

But 2n-1 is odd so (-1)^(2n-1) = -1.

Thus if x=-1 then (-1)^(n-1)x^n /n = -1/n

But this is just a multiple of the harmonic series which diverges.

See: http://en.wikipedia.org/wiki/Harmonic_se…

On the other hand, if x=1 then (-1)^(n-1)x^n/n = (-1)^(n-1) * 1 /n = (-1)^(n-1)/n.

This series converges by the alternating series test. Clearly 1/n -> 0 as n-> infinity and the sequence a_n = 1/n is monotone decreasing.

Check example 1 here:

http://tutorial.math.lamar.edu/Classes/C…