<s>[INST] Convergence of $\sum_{n=1}^{\infty}\frac{1+\sin^{2}(n)}{3^n}$? 
Using either the Direct or Limit Comparison Tests, determine if $\sum_{n=1}^{\infty}\frac{1+\sin^{2}(n)}{3^n}$ is convergent or divergent.


I seem to be completely stuck here.
I've chosen my series to be $\sum\frac{1}{3^n}$, which is clearly a convergent geometric series. But when I do either of the two tests, I get inconclusive answers.
\begin{align}
\lim_{n\to\infty}\frac{1+\sin^2(n)}{3^n}\times\frac{3^n}{1}&=\lim_{n\to\infty}1+\sin^2(n)\\
&=\infty
\end{align}
and
\begin{align}
\frac{1+\sin^2(n)}{3^n}\leq\frac{1}{3^n}\Longrightarrow 1+\sin^2(n)\leq 1
\end{align}
but this inequality is not true.
But I just had a thought now... Am I allowed to choose the series $\sum\frac{2^n}{3^n}$ to help solve this? Clearly $1+\sin^2(n)\leq 2^n,\forall n\geq 1$, and $\sum\frac{2^n}{3^n}$ is also a convergent geometric series.
 [/INST] Since $1\le1+\sin^2(n)\le2$, we can compare to the geometric series
$$
\sum_{n=1}^\infty\frac1{3^n}=\frac12\quad\text{and}\quad\sum_{n=1}^\infty\frac2{3^n}=1
$$
In fact, this is actually the sum of three geometric series:
$$
\begin{align}
\sum_{n=1}^\infty\frac{1+\sin^2(n)}{3^n}
&=\sum_{n=1}^\infty\frac{1-\frac14(e^{2in}-2+e^{-2in})}{3^n}\\
&=\frac32\sum_{n=1}^\infty\frac1{3^n}-\frac12\mathrm{Re}\left(\sum_{n=1}^\infty\frac{e^{2in}}{3^n}\right)\\
&=\frac32\cdot\frac{1/3}{1-1/3}-\frac12\cdot\mathrm{Re}\left(\frac{e^{2i}/3}{1-e^{2i}/3}\right)\\
&=\frac34-\frac12\cdot\mathrm{Re}\left(\frac{e^{2i}/3}{1-e^{2i}/3}\cdot\frac{1-e^{-2i}/3}{1-e^{-2i}/3}\right)\\
&=\frac34-\frac12\cdot\mathrm{Re}\left(\frac{e^{2i}/3-1/9}{10/9-2\cos(2)/3}\right)\\
&=\frac34-\frac14\frac{3\cos(2)-1}{5-3\cos(2)}\\
&=\frac{4-3\cos(2)}{5-3\cos(2)}\\[9pt]
&\doteq0.83996006708282
\end{align}
$$
As estimated, the sum is between $\frac12$ and $1$.
</s>