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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1Did you try taking advantage of the fact that $\lim\limits_{n\rightarrow\infty} \root n\of n=1$? – David Mitra Jan 30 '15 at 20:04
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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. – Meitar Abarbanel Jan 30 '15 at 20:05
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@MeitarAbarbanel $\limsup=\lim$ when the limit exists. – Adam Hughes Jan 30 '15 at 20:07
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I think this series compares with $\sum \frac1{n \log{n}}$. – Ron Gordon Jan 30 '15 at 20:16
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@RonGordon yes, if you show $\sqrt[n]{n}\le M$ ($M=2$ is easy enough) then you can compare with that one. – Adam Hughes Jan 30 '15 at 20:19
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Show that $\sqrt[n]{n}\lt 2$ for all $n\ge 1$. Equivalently show that $n\lt 2^n$, which can be done by induction. – André Nicolas Jan 30 '15 at 20:19
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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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But from some point $n\sqrt[n]{n}$ is strictly smaller then $2n$... – Meitar Abarbanel Jan 30 '15 at 20:11
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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.
Adam Hughes
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Thank you! I have been trying to do this and somehow didn't manage to. – Meitar Abarbanel Jan 30 '15 at 20:09
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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.
Ron Gordon
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