Convergence test on $\sum_{n=2}^\infty\frac{1}{n^2\ln{n}}$

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

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?

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Your limit is actually $1$, not $<1$, which changes matters considerably. Compare with $\sum_{n\ge 2}\frac1{n^2}$. – Brian M. Scott Nov 14 '12 at 7:00

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

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For $n>2$. $\qquad$ – Brian M. Scott Nov 14 '12 at 7:02
In Rudin(Mathematical Analysis), there is a theorem that says: If $a_k$ are non negative and nonincreasing, $\sum a_k$ converges iff $\sum 2^k a_{2^k}$ converges. – Gautam Shenoy Nov 14 '12 at 7:31
@GautamShenoy That would be the Cauchy Condensation test. – Katie Dobbs Nov 14 '12 at 7:32

For the comparison test, just note that $\dfrac{1}{n^2 \log n} < \dfrac{2}{n^2}$ for all $n\geq 2.$

Your application of the ratio test is incorrect as well - just because $a_n < b_n$ holds doesn't mean $\lim a_n < \lim b_n$ , strict equality can hold as well. Try to think of an example. It turns out your limit is actually equal to $1$ so the ratio test is inconclusive.

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Could you please expand on the ratio test? Sorry for asking, but I still need to see all the steps please. – Gineer Nov 14 '12 at 7:06
As you wrote in your post, if the limit in the ratio test is <1, we have converge, if it is >1 then we have diverge, but if it is equal to 1 then we don't know which one. Here, we have the 3rd situation. In your post you got to $n^2 \ln n < (n+1)^2 \ln n+1$, so their ratio is $<1$, but then you incorrectly concluded that the limit of the ratio is $<1.$ Eg if $a_n = n$, $b_n=n+1$ then $a_n < b_n$ but $\lim a_n/b_n =1.$ – Katie Dobbs Nov 14 '12 at 7:10
I don't agree (probably because I'm misunderstanding something ;-): So even when $n\rightarrow\infty$ the numerator would be less that the denominator due to the $+1$ and hence < 1. Are we saying that it is so close that we asume that its $=1$? – Gineer Nov 15 '12 at 7:03
@Gineer We say that $\lim_{n\to\infty} x_n = L$ if for every $\epsilon>0$, there is some $n_0$ such that for all $n>n_0$ we have $| x_n - L|< \epsilon.$ In plain terms: We say the limit of a sequence is $L$ is for any chosen error, even if we pick it to be very small, the sequence eventually remains within our small allowed error. – Katie Dobbs Nov 15 '12 at 14:59
So for the example in my previous comment, the limit is $1$ because for any $\epsilon>0$ you give me, I can find an $n_0$ such that for all $n>n_0$, we have $| a_n/b_n - 1|<\epsilon.$ Indeed, $|a_n/b_n -1| = 1/(n+1)$ so for any given $\epsilon>0$, I say let $n_0 = 1/\epsilon.$ Then certainly for all $n>n_0$ we can see that $|a_n/b_n -1|<\epsilon.$ Thus the limit is 1. – Katie Dobbs Nov 15 '12 at 15:03

For $n>2$ tryto use the integral test over $[2,+\infty)$ in which $f(x)=\frac{1}{x^2\log (x)}$.

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