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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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Do you know how $n^{1/n}$ behaves as $n$ gets big? –  Jonas Meyer Sep 13 '12 at 2:06
    
Well.... actually this is an answer that I DO understand very well (didn't see this comment before). As n gets big that goes to 1, so essentially this summation "converges" to the harmonic series for large n consequently it diverges. –  Squirtle Sep 13 '12 at 2:36
    
dustanalysis: Once you realize that, you can go directly to the limit comparison test: 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 $c>1$. –  Jonas Meyer Sep 13 '12 at 2:42

3 Answers 3

up vote 5 down vote accepted

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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Don't you have to plug in 2^n everywhere then? In particular when taking the (1/n) power you didn't replace that with (1/2^n). Also, I'm not certain which technique you used under the "added" section. I appreciate your help, btw.... but could you add a little more explanation. –  Squirtle Sep 13 '12 at 2:31
    
About your first question: yes, but you're completely right and I oversaw that power. I already fixed it in my post, thanks. About the second proof: it is a standard test for positive series and you can find it here en.wikipedia.org/wiki/Limit_comparison_test –  DonAntonio Sep 13 '12 at 2:49
    
@DonAntonio.... I had a blonde moment though, because it turns out that its obviously true that what you now have is "bigger" than what you had. (I.e. for all x>1, n a positive integer,we have x^(1/2^n) < x^(1/n) therefore, (1/x^(1/2^n)) > (1/x^(1/n)). In other words, the old way I actually think is "better" because the solution looks so clean, thanks for the insightful typo. –  Squirtle Sep 13 '12 at 3:54
    
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 helpful –  DonAntonio Sep 13 '12 at 10:17

HINT: Suppose that you can find a positive number $c$ and a positive integer $n_0$ such that $n^{1/n}<a$ 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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$\displaystyle \sum \frac{2^n}{2^n(2^n)^{1/2^n}}>\sum \frac{2^n}{2^n(2^n)^{1/n}}=\sum\frac{1}{2}=∞$

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After applying Cauchy's Condensation Test –  Squirtle Sep 13 '12 at 4:01

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