# If $\lim_{n\rightarrow\infty} \frac{a_n}{b_n} = 0$ and $\sum_{n=0}^\infty b_n$ is convergent. Proving that $\sum_{n=0}^\infty a_n$ is also convergent

Prove that if $a_n \geq 0, b_n > 0$ for $n \in \mathbb{N}$ , $\sum_{n=0}^\infty b_n$ is convergent and $\lim_{n\rightarrow\infty} \frac{a_n}{b_n} = 0$ then $\sum_{n=0}^\infty a_n$ is also convergent.

What I did was let $\sum_{n=0}^\infty a_n$ be divergent then $\lim_{n\rightarrow\infty} \frac{a_n}{b_n} = 0$ is true if $\lim_{n\rightarrow\infty}b_n = \infty$ which is false, because $\sum_{n=0}^\infty b_n$ is convergent, SO $\lim_{n\rightarrow\infty} a_n$ = 0 which is a contradiction to $\sum_{n=0}^\infty a_n$ being divergent.

My question: Is this correct and if not how can I fix it or solve the problem some other way?

• No, it doesn't look correct. The series $\;\sum a_n\;$ can be divergent and still $\;\lim b_n=0\;$ (example?) – DonAntonio Apr 17 '18 at 18:37
• This is just a special case of the limit comparison test: en.wikipedia.org/wiki/Limit_comparison_test – Umberto P. Apr 17 '18 at 18:39

We cannot assume that $\displaystyle\sum a_{n}$ is convergent.
Rather, $\displaystyle\sum a_{n}=\sum\dfrac{a_{n}}{b_{n}}\cdot b_{n}$, and $\dfrac{a_{n}}{b_{n}}<1$ for large $n$, and hence $\dfrac{a_{n}}{b_{n}}\cdot b_{n}<b_{n}$ and $\displaystyle\sum b_{n}<\infty$.
Exists $\;N\in\Bbb N\;$ such that for all $\;n>N\;$ we have
$$\frac{a_n}{b_n}<1\implies\ldots$$