<s>[INST] 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?
 [/INST] 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.
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