0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ You are using comparison test wrong $\endgroup$ – Yuriy S Oct 23 '18 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 '18 at 10:32
  • $\begingroup$ Can you answer it in more details, please? $\endgroup$ – Mher Khachatryan Oct 23 '18 at 10:34
  • 1
    $\begingroup$ 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 '18 at 10:37
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ You might want to add the (obvious) lower bound $0\le\log(1+x)$. $\endgroup$ – Barry Cipra Oct 23 '18 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 $\endgroup$ – gimusi Oct 23 '18 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. $\endgroup$ – marty cohen Oct 23 '18 at 13:46
  • $\begingroup$ @martycohen Or just use $ \ln(1+x) \sim x$ and limit comparison test to cover both cases. $\endgroup$ – gimusi Oct 23 '18 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$. $\endgroup$ – marty cohen Oct 23 '18 at 13:57
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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