The convergence of $\sum_1^{+\infty}b_n$ follows from the convergence of $\sum_1^{+\infty}a_n$ given that $\frac{a_n}{b_n}\to1$.

This is not true if the series are not eventually constant sign.

Note that $$ \sum_{k=1}^\infty\frac{(-1)^k}{\sqrt{k}} $$ converges, yet $$ \sum_{k=1}^\infty\left(\frac{(-1)^k}{\sqrt{k}}-\frac1k\right) $$ does not converge.

However, $$ \frac{\frac{(-1)^k}{\sqrt{k}}-\frac1k}{\frac{(-1)^k}{\sqrt{k}}} =1-\frac{(-1)^k}{\sqrt{k}} $$ which tends to $1$.