Answer
Divergent
Work Step by Step
Given $$\sum_{n=1}^{\infty} \frac{1+4^{n}}{1+3^{n}}$$
Use the Limit Comparison Test with $a_n =\dfrac{1+4^{n}}{1+3^{n}}$ and $b_n=\dfrac{ 4^{n}}{ 3^{n}}$
\begin{align*}
\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{1+4^{n}}{1+3^{n}}\frac{3^n}{4^n}\\
&=\lim _{n \rightarrow \infty} \frac{1+4^{n}}{4^{n}} \cdot \frac{3^{n}}{1+3^{n}}\\
&=\lim _{n \rightarrow \infty}\left(\frac{1}{4^{n}}+1\right) \cdot\lim _{n \rightarrow \infty}\left( \frac{1}{\frac{1}{3^{n}}+1}\right)\\
&=1
\end{align*}
Since $\displaystyle \sum_{n=1}^{\infty} \frac{ 4^{n}}{ 3^{n}}$ is divergent (geometric series $|r|>1$) , then $\displaystyle\sum_{n=1}^{\infty} \frac{1+4^{n}}{1+3^{n}}$ is also divergent.