sequences and series - Convergence test on $sum_{n=2}^inftyfrac{1}{n^2ln{n}}$

Question
Consider the series $$\sum_{n=2}^\infty\frac{1}{n^2\ln{n}}$$ for each of the following convergence tests, state with justification if the test proves convergence, divergence or confirms neither

• The Ratio Test


• The Comparison Test

My attempt at an Answer
The Ratio test states that a series is:
- absolutely convergent if $\lim_{n\rightarrow\infty}\frac{\lvert u_{n+1}\rvert}{\lvert u_n\rvert}<1$,
- divergent if $\lim_{n\rightarrow\infty}\frac{\lvert u_{n+1}\rvert}{\lvert u_n\rvert}>1$, and
- undefined if $\lim_{n\rightarrow\infty}\frac{\lvert u_{n+1}\rvert}{\lvert u_n\rvert}=1$

so $$u_n=\frac{1}{n^2\ln{n}}$$ $$u_{n+1}=\frac{1}{(n+1)^2\ln{(n+1)}}$$ $$\lim_{n\rightarrow\infty}\frac{\lvert\frac{1}{(n+1)^2\ln{(n+1)}}\rvert}{\lvert\frac{1}{n^2\ln{n}}\rvert}=\lim_{n\rightarrow\infty}\frac{n^2\ln{(n)}}{(n+1)^2\ln{(n+1)}}$$ but $$n^2\ln{(n)}<(n+1)^2\ln{(n+1)}$$ $$\color{red}{\therefore\lim_{n\rightarrow\infty}\frac{n^2\ln{(n)}}{(n+1)^2\ln{(n+1)}}<1}$$ and so absolutely convergent
but $$\lim_{n\rightarrow\infty}\frac{n^2\ln{(n)}}{(n+1)^2\ln{(n+1)}}=1$$ and so is undefined for this test. $\square$

The comparison test has me stumped though.
How do I break $\frac{1}{n^2\ln{n}}$ into multiple terms to perform the comparison test?

Answer

Try $$\frac{1}{n^2\ln n}<\frac{1}{n^2}.$$