2
$\begingroup$

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?

  • 2
    Your limit is actually $1$, not <1, which changes matters considerably. Compare with $\sum_{n\ge 2}\frac1{n^2}$.2012-11-14

3 Answers 3

4

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

  • 3
    @GautamShenoy That would be the Cauchy Condensation test.2012-11-14
4

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.

  • 0
    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 1.2012-11-15
3

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