Test the convergence of $$ \sum_{n=1}^{\infty}\left(\frac{n^2}{2^n}+\frac{1}{n^2}\right) $$
$$ \frac{u_{n+1}}{u_{n}}=\frac{\frac{(n+1)^2}{2^{n+1}}+\frac{1}{(n+1)^2}}{\frac{n^2}{2^{n}}+\frac{1}{n^2}} $$
$$ \lim_{n\to\infty}\frac{u_{n+1}}{u_{n}}=\lim_{n\to\infty}\frac{\frac{(n+1)^2}{2^{n+1}}+\frac{1}{(n+1)^2}}{\frac{n^2}{2^{n}}+\frac{1}{n^2}} $$
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.
A straightforward computation gives $$ \lim_{n\to\infty}\frac{u_{n+1}}{u_{n}} =\lim_{n\to\infty}\frac{\frac{(n+1)^2}{2^{n+1}}+\frac{1}{(n+1)^2}}{\frac{n^2}{2^{n}}+\frac{1}{n^2}} =\lim_{n\to\infty}\frac{(n+1)^4+2^{n+1}}{2^{n+1}(n+1)^2}\frac{n^2 2^n}{n^4+2^n} =\lim_{n\to\infty}\frac{n^2}{(n+1)^2}\frac{2^n}{2^{n+1}}\frac{(n+1)^4+2^{n+1}}{n^4+2^n} =\lim_{n\to\infty}\frac{n^2}{(n+1)^2}\lim_{n\to\infty}\frac{1}{2}\frac{\frac{(n+1)^4}{2^{n+1}}+1}{\frac{n^4}{2^{n+1}}+\frac{1}{2}} =1\cdot\frac{1}{2}\frac{1+1}{0+\frac{1}{2}}=1 $$ Since this limit equals 1 you can't say anything about convergence. But you can apply another approaches. For example cnsider two sums as it was done in Sami Ben Romdhane's answer. I want to emphasize that you can even find exact value for this sum. Indeed, decompose $$ \sum\limits_{n=1}^\infty\left(\frac{n^2}{2^n}+\frac{1}{n^2}\right)=\sum\limits_{n=1}^\infty\frac{n^2}{2^n}+\sum\limits_{n=1}^\infty\frac{1}{n^2} $$ And see this and this questions.
