Is There a Convergence Test for the Series 1/(n*n^(1/n))?

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Homework Help Overview

The discussion revolves around testing the convergence of the series defined by a_[n] = 1/(n*n^(1/n)). The subject area includes series convergence tests and inequalities related to sequences.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply both normal and limit comparison tests, questioning the appropriateness of their comparisons and the resulting inequalities. Some participants raise concerns about the correctness of the inequalities used in the comparisons.

Discussion Status

Participants are actively exploring different comparison strategies and questioning the validity of assumptions made in the inequalities. There is a suggestion that comparing with a constant value may yield better results, indicating a productive direction in the discussion.

Contextual Notes

There is an indication of uncertainty regarding the behavior of n^(1/n) as n approaches larger values, and the original poster expresses difficulty in obtaining a limit for the limit comparison test.

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Homework Statement


Test for convergence the series:
a_[n] = \frac{1}{n*n^{\frac{1}{n}}}


Homework Equations


Various Sequence Convergence Tests


The Attempt at a Solution


So far I've tried both a normal comparison and limit comparison test with n^2. The normal one seemed fine until the end. Here was my logic:

For n greater than 1 (its just less than, not equal)

n^{\frac{1}{n}} \le n

\frac{1}{n^{\frac{1}{n}}} \ge \frac{1}{n}

\frac{1}{n*n^{\frac{1}{n}}} \ge \frac{1}{n^2}}

But that doesn't work because it just proves that for every term, this sequence is greater than the p-series for n^2.

For the limit comparison test I don't get an actual limit, so I can't use it.

If anyone has any suggestions for which test to use, or what series to compare it to, I would be most grateful.
 
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Would n^{\frac{1}{n}} \le \log n work?
 
uh, I'm pretty sure that your inequality is backwards
 
Piamedes said:
uh, I'm pretty sure that your inequality is backwards

I'm pretty sure I'm right for "large" n. And on further reflection, n^(1/n) < 2 is even better.
 
thanks, the comparison with 2 works perfectly
 

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