Determine Convergence: $\sum_{0}^{\infty}\frac{n\ln{n}}{3^n}$

Question

Determine convergence, absolute convergence, or conditional convergence: $\sum_{0}^{\infty}\frac{n\ln{n}}{3^n}$

So, I'm having some trouble with the problem above and I want to know if this is the correct answer using Divergence Test (if $\lim_{n\to\infty}!=0$, then diverges).

(1) So If is LHospitals rule as it is $\frac{\infty}{\infty}$, I get $\frac{1 + \ln{n}}{3^n\ln{n}}$, Which if I take the derivative of again, I would find to converge to 0, so thus it should converge as it isn't harmonic, but Divergence Test doesn't determine convergence, so to be absolutely sure I need to use another test.

(2) So the next test I was thinking about using is the Direct Comparison Theorem. Where, $\frac{1}{3^n}<\frac{n\ln{n}}{3^n}$. Of which, $\frac{1}{3^n}$ converges by geometric series, thus because it's smaller and converges, I can't assume anything.

So now, I'm wondering if I made a mistake anywhere or if there is another test that I can use that is applicable, any help is appreciated. Thanks!

Answer 1

Hint $:$ Let $a_n = \frac{n\ln{n}}{3^n},$ $n \geq 1.$ Observe that $$\lim\limits_{n \rightarrow \infty} \frac {a_{n+1}} {a_{n}} = \frac 1 3<1.$$

Answer 2

The hint:

Take $$\lim_{n\rightarrow+\infty}\frac{a_{n+1}}{a_n}.$$

Answer 3

Note that$$\frac{n\ln n}{3^n}<\frac1{2^n}\iff n\ln n<\left(\frac32\right)^n$$and the last inequality holds if $n$ is large enough, because $\ln n$ is always smaller than $n$. So, $n\ln n<n^2$ and $n^2<\left(\frac32\right)^n$ if $n$ is large enough, because an exponential (with base greater than $1$) always growths faster than any polynomial function. So, by the comparison test, your series converges absolutely.