$$\sum_{n=1}^\infty = \frac{(n^{2/3})}{1+2n}$$

Determine whether the series is divergent or convergent, and if convergent, then whether absolutely or conditionally convergent:

I used the ratio test to determine this answer, But my result is that the limit is simply 1. meaning it is inconclusive. I would just like to know if that is right or perhaps I made a mistake somewhere. thank you.

  • $\begingroup$ The ration test is inconclusive; try comparison. $\endgroup$ – Angina Seng Jun 21 '17 at 17:23
  • $\begingroup$ ohhh, okay I will go try that, thank you $\endgroup$ – M.Bucciacchio Jun 21 '17 at 17:25

$$\sum \frac{n^{2/3}}{2n+1}\geq\sum\frac{1}{2n+1}$$

Edit: The series on the right is divergent. Hence, by Basic Comparison Test so is the original series. If you wish to do this using limit comparison test then

Let $$a_n= \frac{n^{2/3}}{2n+1}=\frac{n^{2/3}}{n\left(2+\frac1n\right)}=\frac{1}{n^{1/3}\left(2+\frac1n\right)}$$

Let$$ b_n=\frac{1}{n^{1/3}}$$

Then $$\lim\frac{a_n}{b_n}=\frac{1}{\left(2+\frac1n\right)}=\frac12$$ which is finite and non-zero. Hence, by Limit Comparison Test, $\sum a_n$ diverges. (since $\sum b_n$ diverges)

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  • $\begingroup$ okay, so if i understand the limit comparison test, i can take the first series and divide it by the second series ? $\endgroup$ – M.Bucciacchio Jun 21 '17 at 17:51
  • $\begingroup$ @M.Bucciacchio I have edited my answer to give more details. Let me know if you have trouble anywhere. $\endgroup$ – Sahiba Arora Jun 21 '17 at 18:02
  • $\begingroup$ thank you so much this cleared it up nicely, i just find comparison tests very confusing. $\endgroup$ – M.Bucciacchio Jun 21 '17 at 18:12
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    $\begingroup$ @M.Bucciacchio Practice. Practice. Practice. $\endgroup$ – Sahiba Arora Jun 21 '17 at 18:16

Note that $$ \frac{n^{2/3}}{1+2n}\sim\frac{1}{n^{1/3}} $$ so that by the limit comparison test and the p-series test the series diverges ($1/3<1$).

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