Find whether the series $\sum_{n = 1}^{ \infty }\dfrac{ n^2 }{ 2^n } $ is absolutely convergent, conditionally convergent, or divergent. Use the ratio test.
My reasoning is as follows.
I use the ratio test: $$\dfrac{ \frac{(n+1)^2}{2^{n+1}} }{ \frac{n^2}{2^n} } = \dfrac{2^n(n+1)^2}{(2^{n+1})n^2}= \dfrac{(n+1)^2}{2n^2}$$ then $$\lim_{n \to \infty} \begin{vmatrix}{ \dfrac{(n+1)^2}{2n^2}} \end{vmatrix} = \infty$$
Therefore, $\sum_{n = 1}^{ \infty }\dfrac{ n^2 }{ 2^n } $ diverges.
However, I am told that the limit actually equals $\dfrac{1}{2}$ as $n \to \infty$. Where is my reasoning incorrect, why is it incorrect, and what is the correct reasoning?
Thank you.