$a_n = \frac{1}{n}b_n$, $\lim b_n = L>0, L\in\mathbb{R}$, prove $\sum a_n$ diverges I have to prove that if
$$a_n = \frac{1}{n}b_n$$for $n\ge 1$ and $$\lim_{n\to\infty} b_n = L>0, L\in\mathbb{R}$$ then $$\sum_{n=1}^{\infty} a_n$$ diverges.
My idea was to show that it's not true that $a_n\to 0$ but I guess it's true because in $\frac{1}{n}b_n$, $b_n$ is limited because converges, and $\frac{1}{n}$ goes to $0$, and there is a theorem that says that when these two things happen in a product, it goes to $0$. So I cannot affirmate anything with this result. I guess it has something to do with comparsion but I cannot find any good comparsion between $a_n$ and $\frac{1}{n}b_n$ 

By the second comparison test, looking at $\sum_k a_k$ and $\sum_k b_k$ if $a_k \sim b_k$ then either both diverge or both converge. Now clearly for $n>n_0 \ b_n$ is within a certain small radius of L. Then you can use the comparison test to establish divergence.
