Recurrence with multiplication Let $\{a_{n}\}$ be a sequence of nonnegative numbers such that $a_{n} = 2^{n}a_{n - 1}^{3/2}$. If $a_{1}$ is sufficiently small, why must $a_{n} \rightarrow 0$ as $n \rightarrow \infty$?
 A: If $a_{n_0}=0$ for some positive integer $n_0$ then $a_n=0$ for every $n\geq n_0$ and we are done.
So let us suppose that $a_n>0$ for every $n$. Let $b_n=\ln(a_n)$. We have
$$
b_n=n\ln2+\frac{3}{2}b_{n-1}
$$
So
$$
\left(\frac{2}{3}\right)^nb_n=\left(\frac{2}{3}\right)^{n-1}b_{n-1}+ (\ln2)n\left(\frac{2}{3}\right)^n
$$
So, adding from $n=2$ to $n=m$ and cancelling $(2/3)$ we get
$$
\left(\frac{2}{3}\right)^{m-1}b_m=  b_{1}+ (\ln2)\sum_{k=2}^{m}k\left(\frac{2}{3}\right)^{k-1}
$$
But $$\sum_{k=2}^{m}k\left(\frac{2}{3}\right)^{k-1}=8+2(3+n)\left(\frac{2}{3}\right)^{m-1}$$
Thus
$$
b_m=\left(\frac{3}{2}\right)^{m-1}\ln(2^8a_1)+ (\ln4)(3+m)
$$
So, if $2^8a_1<1\iff a_1<\dfrac{1}{256}$, we have $\ln(2^8a_1)<0$ and $\lim\limits_{m\to\infty}b_m=-\infty$, therefore $\lim\limits_{m\to\infty}a_m=0$.$\qquad\square$
A: It can be done with a little less effort. First assume that $0 < a_1 < \dfrac{1}{2}$, and calculating several terms of the sequences, we are led to the conclusion that: 
$0 < a_n < 2^{100n}\cdot a_1^{(\frac{3}{2})^{n-1}} < 2^{100n}\cdot \left(\dfrac{1}{2}\right)^{(\frac{3}{2})^{n-1}} = 2^{100n - \left(\frac{3}{2}\right)^{n-1}} = 2^{b_n}$.
It is easy to see that $\displaystyle \lim_{n \to \infty} b_n = \displaystyle \lim_{n \to \infty} 100n - \left(\dfrac{3}{2}\right)^{n-1} = -\infty$.
Thus: $\displaystyle \lim_{n \to \infty} 2^{b_n} = 0$. 
Thus by squeeze theorem: $\displaystyle \lim_{n \to \infty} a_n = 0$
