Can a Convergent Series Prove the Divergence of Another Series?

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SUMMARY

The discussion centers on proving that if the series ∑_(n=1)^∞ a_n is convergent, then the series ∑_(n=1)^∞ 1/a_n is divergent. It is established that since lim n→∞ a_n = 0 for a convergent series, there exists an N such that for all n > N, a_n > 1. Consequently, by the comparison test, ∑_(n=1)^∞ 1/a_n diverges, as the terms 1/a_n exceed 1 for sufficiently large n.

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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.
 

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