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How to check if this series converges or diverges? $$\sum_{n=1}^{\infty} \ln\left({1+ \frac{1}{n\sqrt[3]{n}}}\right)$$ I've tried using comparison test so. $$c_n=\ln\left(\frac{1}{n^{4/3}}\right) <\ln\left({1+ \frac{1}{n\sqrt[3]{n}}}\right)<\ln\left(1+\frac{1}{n}\right)=b_n$$ By integral test $c_n$ and $a_n$ is divergent, so initial series should be divergent too. But using wolfram mathematica it shows that series is convergent. Any ideas?

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  • $\begingroup$ You are using comparison test wrong $\endgroup$
    – Yuriy S
    Oct 23, 2018 at 10:30
  • $\begingroup$ In this case it's better to use the Maclaurin series for $\log(1+x)$. Or just compare to $\sum_{n=1}^\infty \frac{1}{n^{4/3}}$. $\endgroup$
    – Yuriy S
    Oct 23, 2018 at 10:32
  • $\begingroup$ Can you answer it in more details, please? $\endgroup$ Oct 23, 2018 at 10:34
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    $\begingroup$ For better understanding, please read again the definition and conditions for the comparison test and see how to do it right. I'm not going to provide the full solution, but I'm sure someone else will $\endgroup$
    – Yuriy S
    Oct 23, 2018 at 10:37

2 Answers 2

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Recall that for any $x\ge0 \implies \log(1+x)\le x$, therefore

$$0\le \ln\left({1+ \frac{1}{n\sqrt[3]{n}}}\right)\le \frac{1}{n\sqrt[3]{n}}$$

and since $\sum \frac{1}{n\sqrt[3]{n}}$ converges by $p$ test also the given series converges by comparison test.

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  • $\begingroup$ You might want to add the (obvious) lower bound $0\le\log(1+x)$. $\endgroup$ Oct 23, 2018 at 13:27
  • $\begingroup$ @BarryCipra Yes that was implicitely assumed since $1+ \frac{1}{n\sqrt[3]{n}}\ge 1$, thanks for your kind suggestion. Bye $\endgroup$
    – user
    Oct 23, 2018 at 13:30
  • $\begingroup$ A lower bound that can be used to prove divergence (when the sum actually divergences, not this case) is $\ln(1+x) > x/2$ for small enough $x$. Probably $0 < x < 1/2 $ works. $\endgroup$ Oct 23, 2018 at 13:46
  • $\begingroup$ @martycohen Or just use $ \ln(1+x) \sim x$ and limit comparison test to cover both cases. $\endgroup$
    – user
    Oct 23, 2018 at 13:49
  • $\begingroup$ You don't need limits. By the alternating series result, $x-x^2/2 < \ln(1+x) < x$ for $ 0<x<1$. $\endgroup$ Oct 23, 2018 at 13:57
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Since$$\lim_{x\to0}\frac{\log(1+x)}x=\log'1=1,$$you have$$\lim_{n\to\infty}\frac{\log\left(1+\frac1{n\sqrt[3]n}\right)}{\frac1{n\sqrt[3]n}}=1.$$So, since the series $\sum_{n=1}^\infty\frac1{n\sqrt[3]n}=\sum_{n=1}^\infty n^{-\frac43}$ converges, your series converges too.

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