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