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Suppose that sum of $\{a_n\}$ converges and that $a_n > 0$ for all $n ∈ \mathbb{N}$. Prove that sum of $\{1/a_n\}$ diverges.

I tried using contradiction but couldn't come to any conclusion. I also tried using the fact that $\sum a_n$ converges $\implies$ $\lim a_n = 0$.

Thanks in advance for any help! :)

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    $\begingroup$ Those are all good things to try. For instance, if $\lim_{n\to\infty} a_n = 0$, can it also be true that $\lim_{n\to\infty} \frac{1}{a_n} = 0$? $\endgroup$ – Matthew Leingang Mar 20 '17 at 15:59
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If the sum of $\{a_n\}$ converges then by the divergence test, $\lim\limits_{n\to\infty}a_n = 0$. This means that (with limit definitions), $\exists N\in\mathbb{Z^+}$ such that $\forall n>N$, we have $|a_n| < 1$. Thus, $\forall n>N$, we have $|\frac{1}{a_n}|>1$. Ergo, $\lim\limits_{n\to\infty}|\frac{1}{a_n}|>1$. Regardless of the sign, its a given that $\lim\limits_{n\to\infty}\frac{1}{a_n}\ne 0$, and so the sum of $\{\frac{1}{a_n}\}$ diverges.

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$(\sum_{j=1}^n a_j)(\sum_{k=1}^n \frac1{a_k}) \ge n $ because you always have the $n$ terms with $j=k$.

Therefore if $\sum_{j=1}^n a_j < M $, then $\sum_{k=1}^n \frac1{a_k} \ge \dfrac{n}{\sum_{j=1}^n a_j} \gt \dfrac{n}{M} $ which diverges.

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