The problem is this:
Show that if $a_n > 0$ and $\lim_{n\to \infty} na_n = L$ with $L \neq 0$, then the series $\sum a_n$ diverges. (from Abbott's Understanding Analysis, p. 68).
I want someone to correct my proof if it is wrong.
My proof:
Since $(\forall \epsilon >0)(\exists N\in \mathbb N)(\forall n\ge N)|na_n - L|＜\epsilon$ is given, $(\forall n\ge N) L-\epsilon_0 < na_n < L+\epsilon_0$.
And assumed $N \neq 0$, $(\forall n \ge N) \frac{L-\epsilon_0}{n} < a_n < \frac{L+\epsilon_0}{n}$.
Then, let $\epsilon_0 = L - 1$. Then, we get $1/n < a_n < (2L-1)/n$.
Since $1/n < a_n$, by Comparison Test, $\sum a_n$ diverges.
Since the solution from the author is different from mine, I suspect whether there is a wrong step. I wasn't fully confident with defining e_0, but I have no idea whether it is correct or not.
Thank you!