Answer
$\displaystyle\frac{9^n}{3+10^n}$ is convergent
Work Step by Step
Let \[\sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\frac{9^n}{3+10^n}\]
\[\Rightarrow a_n=\frac{9^n}{3+10^n}\]
\[\frac{9^n}{3+10^n}\leq\frac{9^n}{10^n}=\left(\frac{9}{10}\right)^n\]
Also $\displaystyle\sum_{n=1}^{\infty}\left(\frac{9}{10}\right)^n$ is geometric series with common ratio $\displaystyle\frac{9}{10}(<1)$ so $\displaystyle\sum_{n=1}^{\infty}\left(\frac{9}{10}\right)^n$ is convergent.
By using Comparison Test, $\displaystyle\frac{9^n}{3+10^n}$ is convergent.
Hence, $\displaystyle\frac{9^n}{3+10^n}$ is convergent.