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$\begingroup$

I would like to show that

$\sum \frac{1}{nn^{1/n}}$ diverges, and I'm quite certain I will have to use the comparison test. I don't immediately see how that would be useful, though.

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    dustanalysis: Once you realize that, you can go directly to the limit comparison test: http://en.wikipedia.org/wiki/Limit_comparison_test which is what DonAntonio alludes to in his second method. But Brian indicates how you can use that limit together with the direct comparison test, by comparing to $\frac{1}{cn}$ instead of $\frac{1}{n}$, with c>1.2012-09-13

3 Answers 3

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$\displaystyle \sum \frac{2^n}{2^n(2^n)^{1/2^n}}>\sum \frac{2^n}{2^n(2^n)^{1/n}}=\sum\frac{1}{2}=∞$

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    After applying Cauchy's Condensation Test2012-09-13
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HINT: Suppose that you can find a positive number $c$ and a positive integer $n_0$ such that $n^{1/n} for all $n\ge n_0$; then you’d have $\frac1{nn^{1/n}}\ge\frac1{cn}$ for all $n\ge n_0$. What do you know about $\lim_{n\to\infty}n^{1/n}\;?$

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According to Cauchy's Condensation Test , our series converges iff the following series converges:

$\sum_{n=1}^\infty\frac{2^n}{2^n(2^n)^{1/2^n}}=\sum_{n=1}^\infty\frac{1}{2^{n/2^n}}$

But

$2^{n/2^n}=e^{\frac{n}{2^n}\log 2}\xrightarrow [n\to\infty]{}1\neq 0$

thus our series diverges.

Added: Another easy test: choose $\,\{b_n:=\frac{1}{n}\}\,$ ,then:

$\frac{\frac{1}{n\,n^{1/n}}}{b_n}=\frac{1}{\sqrt[n] n}\xrightarrow [n\to\infty]{}1\neq 0$

Thus, our series converges iff the series $\,\sum b_n\,$ does...which it doesn't as it is the harmonic series.

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    Yes @dustanalysis, I know, yet it should be what it must be, and the factor $\,2^n\,$ must appear in that exponent. Good you found it helpful2012-09-13