Finding the interval of convergence I need to find the interval of convergence of the power series $$\sum_{n=1}^\infty (-1)^n\frac {{n^2}{x^n}}{2^n}$$
I have deduced that the radius of convergence is $1$, but am unsure of how to then evaluate the series at $x = -1$ and $x = 1$ to find the interval of convergence. Any help is appreciated.

That's not quite correct: The radius of convergence is $2$, not $1$. To see this, apply the ratio test:
\begin{align*}
\left|\frac{a_{n + 1}}{a_n}\right| &= \frac{(n + 1)^2 x^{n + 1} / 2^{n + 1}}{n^2 x^n / 2^n} \\
&= \frac{(n + 1)^2 x}{2n^2} \\
&= \left(\frac{n + 1}{n}\right) \frac{x}{2}
\end{align*}
Now take a limit as $n \to \infty$.

Now if $x = \pm 2$, our series is
$$\sum_{n = 1}^{\infty} (-1)^n \frac{n^2(\pm 2)^n}{2^n} = \sum_{n = 1}^{\infty} (\mp1)^n n^2$$
The sequence terms do not converge to $0$, and the sum is divergent in either case.
