Showing a series is convergent I am trying to show 
$$ \sum_{n \geq 0} \frac{ 3^n }{\sqrt{2^n + 10^n}} $$
converges. I try to compare as follows 
$$ \frac{ 3^n }{\sqrt{2^n + 10^n}}  < \frac{ 3^n }{\sqrt{10^n} } = \frac{3^n}{ \sqrt{10}^n } = \left( \frac{ 3 }{ \sqrt{10} } \right)^n $$
Since $\sqrt{10} > 3$, then the $\sum \left( \frac{ 3 }{ \sqrt{10} } \right)^n  $ converges and thus the result follows by the comparison test. Is there another method we could have used to show convergence?

There are other ways forward.  Let's invoke the ratio test on the summand $a_n=\frac{3^n}{\sqrt{2^n+10^n}}$.  Proceeding we have
$$\begin{align}
\frac{a_{n+1}}{a_n}&=\frac{3^{n+1}}{\sqrt{2^{n+1}+10^{n+1}}}\,\frac{\sqrt{2^n+10^n}}{3^n}\\\\
&=3\sqrt{\frac{2^n+10^n}{2^{n+1}+10^{n+1}}}\\\\
&=\frac{3}{\sqrt{10}}\sqrt{\frac{1+\left(\frac15\right)^n}{1+\left(\frac15\right)^{n+1}}}\\\\
&\to \frac{3}{\sqrt{10}}\,\,\text{as}\,\,n\to \infty\\\\
&<1
\end{align}$$
Therefore, the ratio test guarantees that the series converges.
We can use the root test with similar success.  Proceeding here, we have
$$\begin{align}
\sqrt[n]{|a_n|}&=\frac{3}{\sqrt{10}}\sqrt[n]{\frac{1}{\sqrt{1+\left(\frac15\right)^n}}}\\\\
&\to \frac{3}{\sqrt{10}}\,\,\text{as}\,\,n\to \infty\\\\
&<1
\end{align}$$
Therefore, the root test guarantees that the series converges.
