Why is the series \(\sum \frac{-1}{n}\) divergent like the harmonic series?

haha1234
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Homework Statement



I know that the harmonic series is divergent.But why \frac{-1}{n} is also divergent?
I've search for some test to test that, but I could not find a method for negative series.
So how can I prove the series is divergent?:confused:

Homework Equations





The Attempt at a Solution

 
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haha1234 said:

Homework Statement



I know that the harmonic series is divergent.But why \frac{-1}{n} is also divergent?
I've search for some test to test that, but I could not find a method for negative series.
So how can I prove the series is divergent?:confused:

Homework Equations



The Attempt at a Solution

\displaystyle (-1)\cdot\sum_{n=1}^\infty \left(\frac{1}{u}\right)=\sum_{n=1}^\infty \left(-\frac{1}{u}\right)
 
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haha1234 said:

Homework Statement



I know that the harmonic series is divergent.But why \frac{-1}{n} is also divergent?
I've search for some test to test that, but I could not find a method for negative series.
So how can I prove the series is divergent?:confused:

Homework Equations





The Attempt at a Solution


If the partial sums of ##\sum\frac 1 n## diverge, then so do the partial sums of ##\sum -\frac 1 n=-\sum\frac 1 n##.
 
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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