Show that if $a_n > 0$ and $lim(na_n) = l$ with $l \not= 0$, then the series $\sum a_n$ diverges

Question

Show that if $a_n > 0$ and $lim(na_n) = l$ with $l \not= 0$, then the series $\sum a_n$ diverges.

$|na_n-l|< \epsilon$, $S_n=a_1+a_2+...+a_n$, suppose $\sum a_n$ converge then $S_n$ converge. So, $S_n$ is cauchy. $|S_n-S_{n-1}| < \epsilon $ => $|nS_n-nS_{n-1}| < \epsilon$, then $|na_n|<\epsilon$, but $l\not=0$, hence contradiction

Are my thoughts correct?

Answer 1

W.L.G,we suppose $l>0$. From $\lim_{n\to\infty} na_n=l>l/2>0$, we know there exists $N$ such that when $n\geq N$ $$na_n>l/2,$$ that is to say when $n\geq N$ $$a_n>\frac{l}{2n},$$ by the comparative method of positive series, the series $\sum a_n$ diverges!

Answer 2

In your proposed solution, when you multiply $\vert S_n - S_{n-1} \vert$ by $n$, $\epsilon$ should also be multiplied by $n$. Hence it doesn't work.

Hint

You have $a_n \ge \frac{l}{2n}$ for $n$ large enough. As the harmonic series diverges, $\sum a_n$ also diverges.