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$$\sum_{n=1}^{\infty}\frac{1}{n^\sqrt{n}}$$ Determine whether this series is convergent or not, with explanation.

Each element is positive, so I've tried bounding it by another convergent series, but couldn't see how.

I couldn't apply integral test, because I couldn't integrate it.

I'm struggling to figure out which convergence/divergence test I should use. (I can use absolute convergence theorems too)

I would really appreciate some help!

Thank you!

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    $\begingroup$ Hint: $\forall n>4$, $n^{\sqrt{n}}>n^2$. $\endgroup$
    – Martigan
    Jun 9, 2015 at 9:21
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    $\begingroup$ Comparing with $\sum \frac{1}{n^p}$ is often one of the simplest ways. $\endgroup$ Jun 9, 2015 at 9:21
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    $\begingroup$ Check this and see whether it answers your questions $\endgroup$
    – creative
    Jun 9, 2015 at 9:23
  • $\begingroup$ how would that help martigan, because according to the theorem in my course (maybe I'm just confused), we would need $$n^\sqrt{n}<n^2$$ $\endgroup$ Jun 9, 2015 at 9:25
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    $\begingroup$ @MatticeVerhoeven Well, you have the inverse of $n^{\sqrt{n}}$... $\endgroup$
    – Martigan
    Jun 9, 2015 at 9:25

2 Answers 2

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Hint: Note that $n^2<n^\sqrt{n}$ for $n>4$. Then compare your series with the one given by the general term $\frac{1}{n^2}$.

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$1\lt \sqrt2\lt\sqrt{n}\,\,\,\,\,\,\,\,\,\,\forall n\ge3\implies0\lt\dfrac{1}{n^{\sqrt{n}}}\lt\dfrac{1}{n^{\sqrt2}}\,\,\,\,\,\,\,\,\,\,\forall n\ge3$

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