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Check whether or not $\sum_{n=1}^{\infty}{1\over n\sqrt[n]{n}}$ converges.

I tried few things but it wouldn't work out. I would appreciate your help.

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    $\begingroup$ Did you try taking advantage of the fact that $\lim\limits_{n\rightarrow\infty} \root n\of n=1$? $\endgroup$ – David Mitra Jan 30 '15 at 20:04
  • $\begingroup$ Shouldn't it be $limsup$? And yes I tried, but then I thought that $lim$ don't take the supremum but what it gets when $n$ tends to infinity. $\endgroup$ – Meitar Abarbanel Jan 30 '15 at 20:05
  • $\begingroup$ @MeitarAbarbanel $\limsup=\lim$ when the limit exists. $\endgroup$ – Adam Hughes Jan 30 '15 at 20:07
  • $\begingroup$ I think this series compares with $\sum \frac1{n \log{n}}$. $\endgroup$ – Ron Gordon Jan 30 '15 at 20:16
  • $\begingroup$ @RonGordon yes, if you show $\sqrt[n]{n}\le M$ ($M=2$ is easy enough) then you can compare with that one. $\endgroup$ – Adam Hughes Jan 30 '15 at 20:19
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$\require{cancel}$

Using the limit comparison test with the harmonic series we see

$$\lim_{n\to\infty} {\cancel{(1/n)}\cdot 1\over \cancel{(1/n)}\sqrt[n]{n}}=1$$

hence the series diverges since the harmonic series does.

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  • $\begingroup$ Thank you! I have been trying to do this and somehow didn't manage to. $\endgroup$ – Meitar Abarbanel Jan 30 '15 at 20:09
  • $\begingroup$ I meant to write the comment under the main question. Shifting. $\endgroup$ – André Nicolas Jan 30 '15 at 20:18
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Since $\lim_{n\to\infty}\sqrt[n]n=1$ then for sufficiently large $n$ we have

$$\frac1{n\sqrt[n]n}\ge \frac1{2n}$$ Can you take it from here?

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  • $\begingroup$ Shouldn't it be $\ge {1\over n}$ from some point? $\endgroup$ – Meitar Abarbanel Jan 30 '15 at 20:08
  • $\begingroup$ Doesn't this hold for all $n\in\mathbb{R^+}$? $\endgroup$ – Lukas Juhrich Jan 30 '15 at 20:09
  • $\begingroup$ @MeitarAbarbanel no, $\sqrt[n]{n}>1$ for all $n\in\Bbb N$. $\endgroup$ – Adam Hughes Jan 30 '15 at 20:10
  • $\begingroup$ But from some point $n\sqrt[n]{n}$ is strictly smaller then $2n$... $\endgroup$ – Meitar Abarbanel Jan 30 '15 at 20:11
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    $\begingroup$ In this case $1$ is sufficiently large;) @Lukas $\endgroup$ – user63181 Jan 30 '15 at 20:12
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$$n \cdot n^{1/n} = n e^{\log{n}/n} = n \left [ 1+ \frac{\log{n}}{n} + O\left ( \frac{\log^2{n}}{n^2} \right ) \right ] = n + \log{n}+ O\left ( \frac{\log^2{n}}{n} \right ) $$

Now,for sufficiently large $n$ (i.e., $n \gt 4$), $n+\log{n} \lt n \log{n} $, so the sum is larger than

$$\sum_{n=2}^{\infty} \frac1{n \log{n}} $$

which diverges.

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