# Does the series diverge or converge [closed]

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

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

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

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• replace $k$ with $k+1$ and then use the integral test – Kenny Lau Aug 9 '18 at 17:46
• "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. – Clement C. Aug 9 '18 at 17:51
• Yet another situation where the comparison test would work if the damn harmonic series diverged like it is supposed to. – The Count Aug 9 '18 at 17:56
• I always recommend to try Cauchy Condensation test when there are logarithms in the series. – Mark Aug 9 '18 at 17:56
• @TheCount How do you compare this to the Harmonic series? – Clement C. Aug 9 '18 at 17:59

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 $\sum \frac 1k$.

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

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

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

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