There are two extensions of the basic comparison test:

Theorem: Suppose that $\displaystyle{\sum_{n=1}^\infty a_n}$ and $\displaystyle{\sum_{n=1}^\infty b_n}$ are positive series; that $c$ is a positive constant; and that $N$ is some positive integer.

If $\displaystyle{\sum_{n=1}^\infty b_n}$ converges and $a_n \le c b_n$ for all $n>N$, then $\displaystyle{\sum_{n=1}^\infty a_n}$ converges.

If $\displaystyle{\sum_{n=1}^\infty b_n}$ diverges and $a_n \ge c b_n$ for all $n>N$, then $\displaystyle{\sum_{n=1}^\infty a_n}$ diverges.

Limit Comparison Test: Let $\displaystyle{\sum_{n=1}^\infty a_n}$ and $\displaystyle{\sum_{n=1}^\infty b_n}$ be positive series. If $$\displaystyle{\lim_{n \to \infty} \frac{a_n}{b_n}}=c,$$
for some positive number $c$, then either both series converge or both diverge.

Example: To determine whether the series $\displaystyle\sum_{n=1}^\infty
\frac{4^n}{2^n+3^n}$ converges or diverges, we'll look for a series that "behaves like" it when $n$ is large. Since
$$
\frac{4^n}{2^n+3^n}\approx \frac{4^n}{3^n},
$$
when $n$ is large, we'll use $\displaystyle\sum_{n=1}^\infty\left(\frac43\right)^n$ for comparison. Since
$$
\lim_{n\to\infty}\frac{ \frac{4^n}{2^n+3^n}}{\frac{4^n}{3^n}}=
\lim_{n\to\infty}\frac{ 3^n}{2^n+3^n}=
\lim_{n\to\infty}\frac{ 3^n}{3^n\left(\frac{2^n}{3^n}+1\right)}=
1>0,
$$
and the geometric series $\displaystyle\sum_{n=1}^\infty\left(\frac43\right)^n$ diverges, we can conclude that our original series diverges as well.