Can a Convergent Series Prove the Divergence of Another Series?

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Homework Statement



Suppose that ∑_(n=1)^∞▒a_n where a_n≠0 is known to be a convergent series. Prove that ∑_(n=1)^∞▒1/a_n is a divergent series.

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The Attempt at a Solution



if ∑_(n=1)^∞▒a_n is convergent then lim n→∞ a_n = 0. thus for some N a_n>1 and a_n >1/a_n if n>N. Thus, ∑_(n=1)^∞▒1/a_n diverges by comparison test.
I don't know if this is right, I don't know what to do.
 

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A series can only converge if it's nth term approaches 0. You see this by looking at the limit of the partial sums. If limit a_n=0 then limit 1/a_n cannot be zero. As I think you were trying to say, for some N>0 and all n>N |a_n|<1. That would mean |1/a_n|>1.
 
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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