Is the series of c(1/2k) divergent?

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Homework Statement



Show \sum_{k=1}^{\infty} c\k^(1\div2k, c is an element of \Re, c &gt; 0, is divergent.<br /> <br /> <h2>Homework Equations</h2><br /> 1/n is divergent<h2>The Attempt at a Solution</h2><br /> Finding a similar series and doing comparison test, is it right?
 
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salbakuta03 said:

Homework Statement



Show \sum_{k=1}^{\infty} c\k^(1\div2k), c is an element of \Re, c > 0, is divergent.

Homework Equations


1/n is divergent


The Attempt at a Solution


Finding a similar series and doing comparison test, is it right?

Assuming the terms in the series are of the form c(1/2k) then the Limit Comparison Test would work too (and probably more cleanly). The series of 1/n is a good choice for this comparison. Your TeX code has the terms as "c\k^(1\div2k)" which renders without any mention to the k after the backslash. Is it displaying incorrectly? What is \k?

--Elucidus

--Elucidus
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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