# Convergence and Divergence of Series with Reciprocal Terms

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.

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

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.
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 $$n$$.