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I am currently studying series and their convergence properties in my calculus course. I came across a problem that I've been trying to solve but haven't been able to crack yet.

Problem Statement:

Suppose that $$\sum_{n=0}^{\infty}a_n$$ is a series of positive terms which is convergent. Show that $$\sum_{n=0}^{\infty}\left(\frac{1}{a_n}\right)$$ is divergent. What about the converse?

My Attempt:

I understand that if a series $$\sum_{n=0}^{\infty}a_n$$ is convergent, then the sequence of its terms {a_n} must approach zero as n approaches infinity. However, I'm not sure how to apply this to the reciprocal series $$\sum_{n=0}^{\infty}\left(\frac{1}{a_n}\right)$$.

For the converse, intuitively it seems that if the reciprocal series is convergent, then the original series should be divergent. But I'm not sure how to prove this formally.

Background:

I am familiar with the basic tests for convergence (like the comparison test, root test, and ratio test), but I'm not sure how to apply them in this case. I would appreciate if the answer could be explained using these or similar concepts.

Thank you in advance for your help!

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    $\begingroup$ If $a_n \rightarrow 0$, then $\tfrac{1}{a_n} \rightarrow \infty$ $\endgroup$ Commented May 6 at 7:17
  • $\begingroup$ Intuitively yes, but how to show it using the series convergence language? $\endgroup$
    – prob1 yuma
    Commented May 6 at 7:27
  • $\begingroup$ If $\lvert a_n \rvert < \varepsilon$, then $\tfrac{1}{\lvert a_n \rvert} >\tfrac{1}{ \varepsilon}$. Maybe this helps $\endgroup$ Commented May 6 at 7:28
  • $\begingroup$ It helps. Many thanks! $\endgroup$
    – prob1 yuma
    Commented May 6 at 9:06

1 Answer 1

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Convergence of $\sum_{n = 0}^\infty a_n$ implies that $a_n \to 0$. Hence $1/a_n \to \infty$ so that $\sum_{n = 0}^\infty 1/a_n$ cannot converge.

The converse

If $\sum_{n = 0}^\infty 1/a_n$ diverges, then $\sum_{n = 0}^\infty a_n$ converges

is not true. As a counterexample take $a_n = 1$ for all $n$.

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