It is true that if $$ \sum_1^{+\infty}a_n\qquad\text{and}\sum_1^{+\infty}b_n $$ satisfies $$ \lim_{n\to+\infty}\frac{a_n}{b_n}=1, $$ then the convergence of $\sum_1^{+\infty}b_n$ follows from the convergence of $\sum_1^{+\infty}a_n$?
What I know is that if there are both positive series, this claim is true.
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 $1$.
The limit comparison test works only for positive $a_n$ and $b_n$ (that is both eventually with constant sign) but it doesn't works in general, for example for the limit case
$$a_n=\frac{1}{ n}$$
$$b_n=\frac{(-1)^n}{ \sqrt n}$$
we have
$$\frac{a_n}{b_n}=\frac{(-1)^n\sqrt n}{n}\to 0 $$
but $a_n$ diverges.
