0
$\begingroup$

$\sum_{k=1}^\infty\frac{1}{k\ln(k+1)}$

I am unable to determine what method to use to test if this series converges or diverges.

My only clue thus far is that there is a similar problem in our text that uses the integral test to determine that the series diverges.

What else is needed here beyond the integral test to determine that the series diverges?

Thanks

$\endgroup$

closed as off-topic by Jyrki Lahtonen, Namaste, Xander Henderson, Taroccoesbrocco, Leucippus Aug 10 '18 at 4:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jyrki Lahtonen, Namaste, Xander Henderson, Taroccoesbrocco, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 4
    $\begingroup$ replace $k$ with $k+1$ and then use the integral test $\endgroup$ – Kenny Lau Aug 9 '18 at 17:46
  • $\begingroup$ "My only clue thus far is that there is a similar problem in our text that uses the integral test to determine that the series diverges." So, why don't you use the integral test, then? It does work. $\endgroup$ – Clement C. Aug 9 '18 at 17:51
  • 1
    $\begingroup$ Yet another situation where the comparison test would work if the damn harmonic series diverged like it is supposed to. $\endgroup$ – The Count Aug 9 '18 at 17:56
  • 1
    $\begingroup$ I always recommend to try Cauchy Condensation test when there are logarithms in the series. $\endgroup$ – Mark Aug 9 '18 at 17:56
  • $\begingroup$ @TheCount How do you compare this to the Harmonic series? $\endgroup$ – Clement C. Aug 9 '18 at 17:59
4
$\begingroup$

Using Cauchy Condensation test we can consider the convergence of the condensed series $\sum 2^k a_{2^k}$ and we have

$$\frac{2^k}{2^k\ln(2^k+1)}=\frac{1}{\ln(2^k+1)}\sim \frac{1}{k\ln 2}$$

which diverges by limit comparison test with $\sum \frac 1k$.

$\endgroup$
  • $\begingroup$ Hi, thanks so much. I found that I was indeed able to use integral/limit comparison to solve as well. $\endgroup$ – jackbenimbo Aug 9 '18 at 18:17
  • $\begingroup$ @jackbenimbo You are welcome! Bye $\endgroup$ – gimusi Aug 9 '18 at 18:22
5
$\begingroup$

$$\sum_{k=1}^\infty\frac{1}{k\ln(k+1)} \ge \sum_{k=1}^\infty\frac{1}{(k+1)\ln(k+1)} = \sum_{k=2}^\infty\frac{1}{k\ln(k)} \ge \int_2^\infty \frac 1 {x \ln x} \ \mathrm dx = [\ln \ln x]_2^\infty = \infty$$

$\endgroup$
1
$\begingroup$

$$\sum_{k=1}^\infty\frac{1}{k\ln(k+1)} \ge \sum_{k=1}^\infty\frac{1}{(k+1)\ln(k+1)}$$ The sum on the right diverges by the integral test.

Hence, by the comparison test, the sum on the left also diverges.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.