I'm trying to prove this sequence converges: $\sum_{n=1}^{\infty} \frac{1}{n\ln^2(n)}$

I noticed that this is continuous function which its derivative is always less than $0$ for $ x \gt 1 $, so I tried to do the integral test but with no success...

Some help?



We can use the integral test for convergence. We have that $$ \sum_{n=3}^\infty\frac1{n\log^2 n}\le\int_2^\infty\frac1{x\log^2 x}\mathrm dx=\biggl[-\frac1{\log x}\biggr]_2^\infty=\frac1{\log 2}. $$

  • $\begingroup$ Can you share how you calculated the integral? $\endgroup$ – FigureItOut Jun 17 '15 at 12:21
  • $\begingroup$ @FigureItOut I used Integral Calculator. $\endgroup$ – Cm7F7Bb Jun 17 '15 at 12:23
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    $\begingroup$ The derivative of $\log(x)$ is $\frac{1}{x}$, and the derivative of $\frac{1}{u}$ is $-\frac{u'}{u²}$. $\endgroup$ – Paul Picard Jun 17 '15 at 12:25

Since the sequence $\dfrac{1}{n\ln^2 n}$ is decreasing, you can apply Cauchy's condensation test, which says that if $a_n$ is decreasing, then $$\sum_{n=1}^\infty a_n <\infty \quad \text{if and only if}\quad \sum_{k=0}^\infty 2^k a_{2^k}<\infty. $$ Typically, this test works well with series with logarithms, you will see why once you apply it.

  • $\begingroup$ thats a nice solution :) $\endgroup$ – supinf Jun 17 '15 at 12:21

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