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Prove that ∑ 1/an diverges if ∑ an converges

Prove that if a_n > 0 for all n \in \mathbb{Z} and if \sum_n a_n converges, then \sum_n \frac{1}{a_n} diverges.


Proof. Since \sum a_n converges we know \lim_{n \to \infty} a_n =0. By the definition limit this means that for all \varepsilon > 0 there exists an integer N such that

    \[ |a_n| < \varepsilon \qquad \text{whenever} \qquad n \geq N. \]

In particular, if we take \varepsilon = 1 then we know there is some N such that n \geq N implies |a_n| < 1. But then,

    \[ |a_n| < 1 \quad \implies \quad \left| \frac{1}{a_n} \right| > 1 \quad \implies \quad \lim_{n \to \infty} \frac{1}{a_n} \neq 0. \]

Hence, the series \sum \frac{1}{a_n} diverges. \qquad \blacksquare

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