<s>[INST] Test the convergence of $\sum_{n=1}^{n=\infty}\left(\frac{n^2}{2^n}+\frac{1}{n^2}\right)$. Test the convergence of 
$$
\sum_{n=1}^{\infty}\left(\frac{n^2}{2^n}+\frac{1}{n^2}\right)
$$
$$
\frac{u_{n+1}}{u_{n}}=\frac{\frac{(n+1)^2}{2^{n+1}}+\frac{1}{(n+1)^2}}{\frac{n^2}{2^{n}}+\frac{1}{n^2}}
$$
$$
\lim_{n\to\infty}\frac{u_{n+1}}{u_{n}}=\lim_{n\to\infty}\frac{\frac{(n+1)^2}{2^{n+1}}+\frac{1}{(n+1)^2}}{\frac{n^2}{2^{n}}+\frac{1}{n^2}}
$$
 [/INST] Note if the two series $\displaystyle \sum_{n=1}^\infty u_n$ and $\displaystyle \sum_{n=1}^\infty v_n$ are convergent then the series $\displaystyle \sum_{n=1}^\infty (u_n+v_n)$ is also convergent.
Now, we know that the Riemann series $\displaystyle \sum_{n=1}^\infty\frac{1}{n^2} $ is convergent and we have 
$$\frac{n^2}{2^n}=_\infty o(\frac{1}{n^2})$$
hence the series $\displaystyle \sum_{n=1}^\infty \frac{n^2}{2^n}$ is also convergent by comparison with the convergent Riemann series.
</s>