# Calculating where X converges

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 ak+1 and an 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 an+1 < an and that the limit as k -> $$\infty$$ = 0, so why does it apparently diverge?

Homework Helper
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 @%$@# for overlooking this. So I assume I just use like a comparison test for 1/ln(k+2) with 1/k (diverges) since ln makes values of k smaller so the frac. gets bigger. Science Advisor Homework Helper Wait so I can't use the alternating series test with a non alternating series ... wow I'm going to kick myself in the @%$@# 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 ak+1 and an 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 $$a_n$$.