1
$\begingroup$

Cauchy ratio test yields 1 (so it's inconclusive). I have tried this:

$$\frac{1}{n \log^2n}=\frac{1}{n \log n \log n}=\frac{1}{\log n^n \log n}\geq \frac{1}{\log n^n -n} \approx \frac{1}{\log n!} $$

Now, since $\sum 1/\log n!$ diverges, the original series must diverge too. But Wolfram Alpha says it's convergent. How did I go wrong and how can I solve this?

  • 1
    Try integral test. $1/n$ is the derivative of $logn$ – 2017-01-27
  • 0
    I haven't seen that one in class, so even if it's useful (and I'm gonna learn it) there must be another way to evaluate this without using that test. – 2017-01-27
  • 0
    On a side note, [here's](https://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/Boas.pdf) a paper that discusses the slow convergence of this series (and others). – 2017-01-27

2 Answers 2

3

You might find it easier to apply the Cauchy condensation test:

$$\sum_{n=2}^\infty\frac1{n\log^2n}\le\sum_{n=1}^\infty\frac{2^n}{2^n\log^22^n}=\sum_{n=1}^\infty\frac1{n^2\log^22}$$

  • 1
    Typo? You mean $\sum_{n=2}^\infty\frac1{n\log^2n}\le\sum_{n=1}^\infty\frac{2^n}{2^n\log^22^n}=\sum_{n=1}^\infty\frac1{n^2\log^22}$ – 2017-01-27
  • 0
    @chilangoincomprendido Oh, nvm, I'm stupid :D Thanks for pointing it out. – 2017-01-27
  • 0
    There is a typo. The argument on the logarithm is $2$, not $n$. – 2017-01-27
  • 0
    This makes more sense. Thanks! – 2017-01-27
  • 0
    @Manuel :-) No problem – 2017-01-27
1

Fundamentally, what went wrong was your inequality chain. We have that $$\frac{1}{\ln \left(x^x\right)-x} \gg \frac{1}{\ln \left(x^x\right)\ln x}$$ contrary to what your inequality chain would imply.

  • 1
    You're right, what a silly mistake :( – 2017-01-27