Convergence of a Series with Alternating Terms: Finding Values of X

[DMOC]

Homework Statement

Determine values for x where \sum \frac{(2^k)(x^k)}{ln(k + 2)} converges if k goes from 0 to infinity.

Homework Equations

Limit test and alternating series test.

The Attempt at a Solution

I used the limit comparison test to test out a_k+1 and a_n and I ended up getting (-1/2) \le x \le (1/2)

Then I had to compare the endpoints. I noticed that the numerator is always 1 in the case of x=1/2 or always 1 or -1 (in the case of x = -1/2).

So by the alternating series test, I concluded that BOTH of the endpoints converges. However, my answer key says that x=1/2 diverges while x=-1/2 converges (I'm correct there).

FYI, for x = 1/2,

I said that a_n+1 < a_n and that the limit as k -> \infty = 0, so why does it apparently diverge?

 

[Dick]

I'm with you at x=(-1/2), but how did you conclude 1/ln(k+2) converges by the alternating series test? It doesn't even alternate.

 

[DMOC]

Wait so I can't use the alternating series test with a non alternating series ... wow I'm going to kick myself in the @%$@# for overlooking this.

So I assume I just use like a comparison test for 1/ln(k+2) with 1/k (diverges) since ln makes values of k smaller so the frac. gets bigger.

 

[Dick]

Wait so I can't use the alternating series test with a non alternating series ... wow I'm going to kick myself in the @%$@# for overlooking this.

So I assume I just use like a comparison test for 1/ln(k+2) with 1/k (diverges) since ln makes values of k smaller so the frac. gets bigger.

Right.

 

[Sweet_GirL]

I used the limit comparison test to test out a_k+1 and a_n and I ended up getting (-1/2) \le x \le (1/2)

This is the Ratio Test not the Limit Comparison Test.
Also, you should write a_k not a_n.