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Calculating where X converges

  • Thread starter DMOC
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  • #1
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Homework Statement



Determine values for x where [tex]\sum \frac{(2^k)(x^k)}{ln(k + 2)}[/tex] 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 [tex](-1/2) \le x \le (1/2)[/tex]

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 -> [tex]\infty[/tex] = 0, so why does it apparently diverge?
 

Answers and Replies

  • #2
Dick
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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.
 
  • #3
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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.
 
  • #4
Dick
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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.
 
  • #5
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I used the limit comparison test to test out ak+1 and an and I ended up getting [tex](-1/2) \le x \le (1/2)[/tex]
This is the Ratio Test not the Limit Comparison Test.
Also, you should write [tex]a_k[/tex] not [tex]a_n[/tex].
 

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