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I need to study the convergence of the series

$\sum_{n=1}^{\infty}(\sqrt[n]{n}-1)$

Now, I know that if we have a series $\sum_{n=1}^{\infty}a_n$ with positives elements and we can find a series $\sum_{n=1}^{\infty}b_n$ so that $0<a_n<b_n$ then if $\sum_{n=1}^{\infty}b_n$ is convergent then $\sum_{n=1}^{\infty}a_n$ is convergent.

Else, if $\sum_{n=1}^{\infty}a_n$ is divergent then $\sum_{n=1}^{\infty}b_n$ is divergent. The problem is I do not really know how to choose the series. Can you help me out?

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  • $\begingroup$ Actually, $\sum_{n=1}^\infty\sqrt[n]n$ diverges. $\endgroup$ – José Carlos Santos Oct 26 '18 at 16:54
  • $\begingroup$ $$\lim_{n\rightarrow \infty}n^{1/n}=1$$ so your comparison sum is divergent... $\endgroup$ – Eleven-Eleven Oct 26 '18 at 16:54
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The series diverges by comparison with $\sum \frac{1}{n}\log n$.

Let $n \ge 2$. Since $\sqrt[n]{n} = e^{\frac{1}{n}\log n}$, the mean value theorem gives $\sqrt[n]{n} - 1 = e^{c_n}\cdot \frac{1}{n}\log n$ for some $c_n\in \left(0, \frac{1}{n}\log n\right)$. Now $e^{c_n} > 1$, so that $$\sqrt[n]{n} - 1 > \frac{1}{n}\log n$$ Now you can compare your series with the divergent series $\sum \frac{1}{n}\log n$.

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  • $\begingroup$ Can you detaliate how you got that $\sqrt[n]{n} - 1 = e^{c_n}\cdot \frac{1}{n}\log n$? For what function you appy mean value theorem? $\endgroup$ – Ghost Oct 26 '18 at 17:14
  • $\begingroup$ @Ghost I apply it to the function $f(x) = e^x$. $\endgroup$ – kobe Oct 26 '18 at 17:15
  • $\begingroup$ So for $f(x)=e^x$ we apply it on the interval $(0, \frac{1}{n}log n)$? $\endgroup$ – Ghost Oct 26 '18 at 17:18
  • $\begingroup$ @Ghost correct. $\endgroup$ – kobe Oct 26 '18 at 17:20
  • $\begingroup$ Now the only problem is how to show that the new series is divergent. $\endgroup$ – Ghost Oct 26 '18 at 17:23
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$$\sqrt[n]n-1={n-1\over(\sqrt[n]n)^{n-1}+(\sqrt[n]n)^{n-2}+\cdots+\sqrt[n]n+1}\ge{n-1\over n+n+\cdots+n+n}={n-1\over n^2}\ge{1\over2n}$$

(with the final inequality assuming $n\gt1$).

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  • $\begingroup$ This is actually alot faster indeed. Nice thinking! But, be careful cause for $n\leq2$ the last inequality is not right. $\endgroup$ – Ghost Oct 26 '18 at 17:31
  • $\begingroup$ (+1) for using elementary analysis only. $\endgroup$ – Mark Viola Oct 26 '18 at 18:25
  • $\begingroup$ @Ghost, thanks for your comment. Note, however, that the final inequality is correct for $n=2$. (My parenthetical remark accounts for the one case, $n=1$, where it's not correct.) $\endgroup$ – Barry Cipra Oct 26 '18 at 19:06
  • $\begingroup$ Yes, you are right! Sorry, I typed it wrong but, for $n<2$ it is not right. Now if we take n as a natural number, the only wrong case is $n=1$ but for n real it is $n<2$. Sorry I did not take into consideration n is natural number. Anyway, great thinking! $\endgroup$ – Ghost Oct 26 '18 at 20:22
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Hint: $$n^{1/n}-1=\left(e^{\log n/n}-1\right)\sim\frac{\log n}{n} $$

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