Another Test for Convergence Question

[Seydlitz]

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

 

[Dick]

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

 

[Jolb]

Edit: Dick beat me to it. Quick.

 

[Office_Shredder]

Try using 3 =e^ln(3) to rewrite your summand

 

[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.

 

[Dick]

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.

 

[Seydlitz]

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!