# Find whether the series $\sum_{n = 1}^{ \infty }\dfrac{ n^2 }{ 2^n }$ is absolutely convergent, conditionally convergent, or divergent. - Ratio test

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.

The limit of $\frac {(n+1)^2} {(2n^2)}$ is $\frac 1 2$, so the series converges, in fact it is $\frac {n^2+2n+1} {2n^2}$, so you put in evidence $n^2$ and it's done! However, it was easier with the root test, not more than a line (same result).
Hint. Note that $$\dfrac{(n+1)^2}{2n^2}=\dfrac{n^2(1+1/n)^2}{2n^2}=\dfrac{(1+1/n)^2}{2}.$$ Then the correct limit easily follows.
$$= \dfrac{2^n(n+1)^2}{(2^{n+1})n^2}=\frac{(n+1)^2}{2n^2}=\frac{(1+\frac{1}{n})^2}{2}$$