Another Test for Convergence Question

1. Sep 10, 2013

Seydlitz

1. The problem statement, all variables and given/known data
Test the following series for convergence or divergence.
$$\sum_{n=1}^{\infty} \frac{1}{3^{\ln n}}$$

3. 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

2. Sep 11, 2013

Dick

Use that 3=e^(ln(3)) and use the laws of exponentials to rearrange it a bit. It becomes a power series.

3. Sep 11, 2013

Jolb

Edit: Dick beat me to it. Quick.

Last edited: Sep 11, 2013
4. Sep 11, 2013

Office_Shredder

Staff Emeritus
Try using 3 =eln(3) to rewrite your summand

5. Sep 11, 2013

Seydlitz

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. Sep 11, 2013

Dick

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

7. Sep 11, 2013

Seydlitz

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