Prove that $\limsup({a_n/b_n)}$ is finite and $\sum {b_n}$ converges, then $\sum {a_n}$ converges

Question

Suppose that ${a_n}\ge 0$ and ${b_n \ge 0}$ for all n. If $\limsup({a_n/b_n)}$ is finite and $\sum {b_n}$ converges, then $\sum {a_n}$ converges.

I spent an hour on this proof, I can't seem to figure it out. Can you guys help me?

Answer 1

Define, $$ \limsup_{n\to\infty}\frac{a_n}{b_n}= \lim_{n\to\infty}\sup_{k\geq n}\frac{a_n}{b_n} = l. $$ Take an $\epsilon>0$. There exists an $N$ such that for all $n \geq N$, we have, $$ \sup_{k\geq n}\frac{a_n}{b_n}\leq l+\epsilon \implies\frac{a_n}{b_n}\leq l+\epsilon,\forall n\geq N. $$ With this in mind, we are almost done. Notice that, \begin{align*} \sum_{n=1}^{\infty}a_n & = \sum_{n=1}^{N-1}a_n + \sum_{n=N}^{\infty}a_n \\ & = \sum_{n=1}^{N-1}a_n + \sum_{n=N}^{\infty}\frac{a_n}{b_n}b_n \\ & \leq \sum_{n=1}^{N-1}a_n + (l+\epsilon)\sum_{n=N}^{\infty}b_n \\ & < \infty. \end{align*}