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Converging or diverging 1/ln(n)

  • #1
Determine if the series n=2 to inf. of 1/ln(n) converges or diverges


Ok so first I tried the limit test (the simple one) and found that it was 0 which was not helpful at all. Then I tried the integral test. It came out to be (integral)1/ln(n)=n/ln(n) + n/(ln(n))^2 + 2(integral from 2 to infin.) 1/(ln(n))^3. I was thinking of possibly doing a direct comparison test, but I have no clue what to compare it to. So then I tried the ratio test. That also failed, because the limit of the absolute value of the ratio was equal to 1 thus inconclusive and leaving me back where I started.

I have no idea how else to approach this problem. I am hoping that I maybe just messed up my integration. If anyone has a clue how to approach this that would be great.
 

Answers and Replies

  • #2
614
0
Try comparing it with the harmonic series
 
  • #3
ahhhh. Just to check. When I do a direct comparison to the harmonic series (which diverges) 1/ln(n) is larger so it must also converge. Is this right or have I been staring at this problem long enough that my logic is swiss cheese?
 
  • #4
614
0
Check the criteria for the comparison test again, you've made one mistake.
 
  • #5
The only criteria I have for the Direct Comparison test is that "if a series is less than or equal to a converging series then it also converges." and "if it is greater than a diverging series it also diverges." Am I missing something?
 
  • #6
o wait....duh since harmonic diverges and the 1/lnn is larger it must diverge. that was stupid on my part
 

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