How do I show that $ k\times0.25^{2^{k+2}-2^k} \to 0$ as $k \to \infty$?

Question

I started by showing that $$ k\times0.25^{2^{k+2}-2^k} =k\frac{0.25^{2^{k+2}}}{0.25^{2^k}}$$

$$= k \frac{({0.25^{2^k}})^4}{{0.25^{2^k}}}$$

$$=k(0.25^{2^k})^3 $$

Then I don't know how to proceed. I suspect the $(0.25^{2^k})^3$ goes to $0$ much faster than $k$ goes to infinity but don't know how to prove it.

Answer 1

Use L'Hospital rule: $$\lim_\limits{k\to\infty} \frac{k}{4^{3\cdot 2^k}}=\lim_\limits{k\to\infty} \frac{1}{3\cdot 4^{3\cdot 2^k}\cdot \ln4 \cdot 2^k \cdot \ln2}=0.$$

Answer 2

Hint: prove that $\,4^{2^k} \ge k^2\,$ for $\,k \ge 0\,$, then squeeze $\displaystyle\,0 \le k \cdot \frac{1}{4^{3 \cdot 2^k}} \le k \cdot \frac{1}{k^2}\,$.