Prove: If $\sum a_n$ converges then $\sum \frac{1}{a_n}$ diverges

Question

Prove: If $\sum a_n$ converges, then $\sum \frac{1}{a_n}$ diverges.

I want to prove this statement. I've been trying to find a way but I couldn't. let's say $a_n$ converges to $a$ then what should i do? Can i prove it like a sequence. For example $\forall\epsilon >0, \exists N>0 \ if \ n>N \ then \ |a-L|< \epsilon$ . I don't think i can apply this to series.

Answer 1

If $\sum a_n$ converges, we must have that $\lim\limits_{n\to\infty}a_n=0$ (otherwise it would have been divergent when we checked the limit by the divergence/limit test). Therefore, if we considered $\sum\frac{1}{a_n}$, when we take the limit, we see that $$\lim_{n\to\infty}\frac{1}{a_n}\left[\to\frac{1}{0}\right]\to\infty$$

Answer 2

Since $\sum a_n$ converges to $s$, say. Then the partial sums $S_n$ converge to a limit $s$. But we have $a_n = S_n - S_{n-1} \to s -s = 0$ by the difference of limits theorem.

So $a_n \to 0 \implies \frac{1}{a_n} \to \infty \neq 0$ so $\sum \frac{1}{a_n}$ can't converge by the contrapositive of $\sum a_n < \infty \implies a_n \to 0$.