For which $a,b,c \in \mathbb R$ series $$\sum_{n=1}^{+\infty} (\arctan (\frac{1}{\sqrt[3]{n}}) + \frac{a}{n}+\frac{b}{n^c} +\frac{c}{n^2})$$ is convergent?

My try:
$$(\frac{1}{\sqrt[3]{n}}) + \frac{a}{n}+\frac{b}{n^c} +\frac{c}{n^2})=\frac{1}{n^{\frac{1}{3}}}+\frac{1}{3n}+ o(\frac{1}{n})+\frac{a}{n}+\frac{b}{n^c} +\frac{c}{n^2}\le 1+ o(\frac{1}{n})+\frac{\frac{1+3a}{3}}{n}+\frac{b}{n^c} +\frac{c}{n^2}$$ Series $\sum(1+ o(\frac{1}{n})+\frac{\frac{1+3a}{3}}{n}+\frac{b}{n^c} +\frac{c}{n^2})$ is convergent when:

  • $\frac{1+3a}{3}=0$
  • $c>1, b\in \mathbb R$ lub $b=0, c\in \mathbb R$
    Then I wanted to tell that from direct comparison test series $\sum_{n=1}^{+\infty} (\arctan (\frac{1}{\sqrt[3]{n}}) + \frac{a}{n}+\frac{b}{n^c} +\frac{c}{n^2})$ is convergent when series $\sum(1+ o(\frac{1}{n})+\frac{\frac{1+3a}{3}}{n}+\frac{b}{n^c} +\frac{c}{n^2})$ is convergent but in direct comparison test I need series $>0$ so I can't use it.

    Can you help me how complete this task?
  • $\begingroup$ @PierreCarre Why $\arctan$ is not problem if when I use Taylor I have $\frac{1}{n^{\frac{1}{3}}}+\frac{1}{3n}+o(\frac{1}{n})$ and $\sum \frac{1}{n^z}$ is convergent for $z>1$ but I have $1/3$ and $1$? $\endgroup$ – MP3129 Apr 17 at 9:56
  • 1
    $\begingroup$ The term $\arctan(1/n^{1/3})$ behaves like $1/n^{1/3}$ and yields a divergent series . Cancelling out this behaviour could be achieved by setting $c=\frac 13$ with some negative $b$, but then you would have to set $a=0$. $\endgroup$ – PierreCarre Apr 17 at 10:02
  • $\begingroup$ @PierreCarre I agree with you, very good idea, but is this the only possible answer? $\endgroup$ – MP3129 Apr 17 at 10:05

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