Test the convergence of $\sum_{n=1}^{n=\infty}\left(\frac{n^2}{2^n}+\frac{1}{n^2}\right)$. Test the convergence of 
$$
\sum_{n=1}^{\infty}\left(\frac{n^2}{2^n}+\frac{1}{n^2}\right)
$$
$$
\frac{u_{n+1}}{u_{n}}=\frac{\frac{(n+1)^2}{2^{n+1}}+\frac{1}{(n+1)^2}}{\frac{n^2}{2^{n}}+\frac{1}{n^2}}
$$
$$
\lim_{n\to\infty}\frac{u_{n+1}}{u_{n}}=\lim_{n\to\infty}\frac{\frac{(n+1)^2}{2^{n+1}}+\frac{1}{(n+1)^2}}{\frac{n^2}{2^{n}}+\frac{1}{n^2}}
$$
 A: 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: 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.
