sequences and series - Using the Root Test to Solve $sum_{k=1}^infty(frac{k^{2}}{2^{k}})$

$\sum_{k=1}^\infty(\frac{k^{2}}{2^{k}})$

So I've taken the $kth$ root of the numerator and denominator and got:

$\sum_{k=1}^\infty(\frac{k^{\frac{2}{k}}}{2})$

$k^{\frac{2}{k}}$ as k approaches infinity. What to do now?

Answer

$$\lim_{k \to \infty} k^{\frac{2}{k}}$$
$$\lim_{k \to \infty} e^{ \frac{2 \ln k}{k}} = 1$$

as the exponent tends to $0$ as $k \to \infty$ by L'Hospital's Rule.

Dividing by two, we get that the limit is $\frac{1}{2} < 1$ and thus the series converges by the root test.