Another Test for Convergence Question

  • Thread starter Seydlitz
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  • #1
Seydlitz
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Homework Statement


Test the following series for convergence or divergence.
$$\sum_{n=1}^{\infty} \frac{1}{3^{\ln n}}$$

The Attempt at a Solution


I've tried to compare this to geometric series ##3^n## but obviously the target term is larger overall than its geometric counterpart. Comparing with the reciprocal of ##\ln n## also brings no avail because the term is smaller than the former. Ratio test gives ##ρ=1## which is inconclusive. It certainly passes preliminary test because the limit goes to 0 once n goes to infinity. I haven't done integral test because it's quite ugly and I don't know how to integrate the corresponding function.

Can you guys give me suggestion to tame this series down nicely?

Thank You
 

Answers and Replies

  • #2
Dick
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Use that 3=e^(ln(3)) and use the laws of exponentials to rearrange it a bit. It becomes a power series.
 
  • #3
Jolb
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Edit: Dick beat me to it. Quick.
 
Last edited:
  • #4
Office_Shredder
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Try using 3 =eln(3) to rewrite your summand
 
  • #5
Seydlitz
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Strange things happened when I arrange the summation.

##3=e^{\ln 3}##
$$3^{\ln n}=(e^{\ln 3})^{\ln n}$$
$$3^{\ln n}=e^{\ln 3 ° \ln n}$$
$$3^{\ln n}=n^{\ln 3}$$

If that's the case then the series must be convergent because of p-test, with p being larger than 1.
 
  • #6
Dick
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Strange things happened when I arrange the summation.

##3=e^{\ln 3}##
$$3^{\ln n}=(e^{\ln 3})^{\ln n}$$
$$3^{\ln n}=e^{\ln 3 ° \ln n}$$
$$3^{\ln n}=n^{\ln 3}$$

If that's the case then the series must be convergent because of p-test, with p being larger than 1.

I don't know if I'd call that strange, but it is the correct conclusion.
 
  • #7
Seydlitz
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I don't know if I'd call that strange, but it is the correct conclusion.

Well maybe I'm not quite used to manipulating exponents yet. Thanks for your help guys!
 

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