0
$\begingroup$

Check the convergence of the infinite series $\sum\limits\frac1{(\log n)^{3/2}}$.

I have tried to use comparison test but got no success.

$\endgroup$
6
  • 2
    $\begingroup$ Hint:$$(\log{(n)})^{3/2}\lt n\qquad\forall n\in\mathbb{N}$$ $\endgroup$ Commented Jul 11, 2020 at 14:00
  • $\begingroup$ What have you tried explicitly? Where are you stuck? See How to ask a good question. $\endgroup$ Commented Jul 11, 2020 at 14:02
  • $\begingroup$ Have you learned the Cauchy Condensation test? That makes the term you're summing over immediately much more manageable. $\endgroup$ Commented Jul 11, 2020 at 14:14
  • $\begingroup$ Thanks for replying 😊. I used the comparison test to prove that 1/log n is divergent (with 1/n). I wasn't able to get such divergent series to make a conclusion. $\endgroup$ Commented Jul 11, 2020 at 14:26
  • $\begingroup$ @ Peter Foreman Thank u for the hint. can you also give a proof for this identity? $\endgroup$ Commented Jul 11, 2020 at 14:29

1 Answer 1

4
$\begingroup$

More generally, For $\sum\limits_{n=2}^\infty \frac{1}{(\operatorname{log}n)^p}$

We use Cauchy Condensation test,

\begin{eqnarray} \sum\limits_{n=2}^\infty 2^n \frac{1}{(\operatorname{log}2^n)^p}=\frac{1}{(\operatorname{log}2)^p} \sum\limits_{n=2}^\infty \frac{2^n}{n^p} \end{eqnarray}

which diverges for all $p>0$. Since $2^nn^{-p}\to \infty$ for all $p>0$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .