Suppose that $\sum a_n$ converges and that $a_n \neq 0$ for all $n \in \mathbb N $. Prove that $\sum \frac{1}{a_n}$ diverges.
My attempt
let the series $\sum a_n$ converge to a. Therefore $\lim_{n\to\infty} S_n = a$
Yeah. I have never tackled a similar question to this. Could anyone help me get started?
Theorem: If $\sum a_i$ converges, then $a_i\to 0$.
It shouldn't be hard to prove that if one of the sequences in question goes to $0$ the other one doesn't. When $x$ gets small, $x^{-1}$ gets big.
By assumption we have $a_{n} \to 0$ (this is a very basic theorem.). Then for every $\varepsilon > 0$ there is some $N$ such that $a_{n} \leq |a_{n}| < \varepsilon$ for all $n \geq N$; so $$ \frac{1}{a_{n}} > \frac{1}{\varepsilon} $$ for all $n \geq N$. This shows that the sequence $(1/a_{n})$ diverges to infinity; hence by the same theorem we employ above the corresponding series diverges.
