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I have been trying to solve this problem for many hours, but I am lost. I would like to use the Comparison Test. However, any other test would be also OK.

I am interested not only in the exponent 0.5. I am interested in all the exponents that are elements of the interval (0,1).

log in the image is natural log.

I am posting "the solution" of WolframAlpha.

Thank you for your help!

enter image description here

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    Is your question that you want a proof about the convergence or divergence of $$\sum_{n=2}^{\infty}\frac{1}{n\log^k(n+1)}$$ for all $k \in (0,1)$? In particular, a proof that uses the comparison test? – 2017-01-31
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    According to WolframAlpha the series diverges. I would prefer a proof that use the Comparison Test. But, any other test is also welcome. – 2017-01-31
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    I used the integral test for a similar series that inside the log has n instead of (n+1). That series diverges. However, I cannot use it for the Comparison Test because that series is greater that the one of the problem. – 2017-01-31
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    Compare with $\sum_{n=2}^\infty\frac{1}{n\ln n}$, this one you can use the integral test to show the divergence. – 2017-01-31
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    As the series is known to diverge, it suffices to show divergence for $k=1$ (the fastest decrease rate). – 2017-01-31
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    @FrankLu I checked and it is working! Great idea thank you so much! – 2017-01-31
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    @YvesDaoust Excellent comment! Thank you! – 2017-01-31
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    @FrankLu One question. What can I do if I need also 1 in the interval? For that point the series that you found does not work because the one of the problem is smaller, then I cannot apply the Comparison Test. – 2017-01-31
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    @YvesDaoust The series that Frank Lu suggested does not work for k=1. May you help me with one that works for k=1? – 2017-01-31
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    If starts at $k=1$ you can try $\sum_{k=1}^\infty\frac{1}{n\ln(n+1)}$. This one also works, but just need to be a little careful to make the argument rigorors. Essentially if you can prove that the tail of this series diverges, you are done. – 2017-01-31

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For $\alpha\in(0,1)$, note that $$\frac{1}{n\ln^\alpha (n+1)}>\frac{1}{n\ln (n+1)}=\frac{1}{n(\ln n+\ln\frac{n+1}{n})}>\frac{1}{2n\ln n},$$ whenever $n$ is sufficiently large, for example say $n>4$. Then by integral test you can show that $\sum_{n=5}^\infty\frac{1}{2n\ln n}$ diverges. It follows that $$\sum_{n=1}^\infty\frac{1}{n\ln^\alpha(n+1)}\geq\sum_{n=1}^4\frac{1}{n\ln^\alpha(n+1)}+\sum_{n=5}^\infty\frac{1}{2n\ln n}=+\infty.$$

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    Amazing proof! Thank you so much! – 2017-02-02
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    You're welcome! – 2017-02-02
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By the Cauchy condensation test, this compares to the following sum:

$$2\sum_{n=2}^\infty\frac1{n\ln^kn}\ge\sum_{n=1}^\infty\frac{2^n}{2^n\ln^k2^n}=\sum_{n=1}^\infty\frac1{n^k\ln^k2}=\frac1{\ln^k2}\sum_{n=1}^\infty\frac1{n^k}$$

Again by the Cauchy condensation test, this compares to the following sum:

$$2\sum_{n=1}^\infty\frac1{n^k}\ge\sum_{n=0}^\infty\frac{2^n}{(2^n)^k}=\sum_{n=0}^\infty(2^{1-k})^n$$

which is a geometric series and diverges when $k\le1$.

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    Fantastic proof! Thank you so much! – 2017-01-31
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    :D Your welcome! – 2017-01-31
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    It is more relevant to say that the series diverges when $k\le1$, isn't it ? – 2017-02-01
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    @YvesDaoust oh, right lol. – 2017-02-01