Proof verification for $\lim_{n\to\infty}\frac{a^n}{(1+a)(1+a^2)\cdots(1+a^n)}$

Question

Find the limit: $$ \lim_{n\to\infty}\frac{a^n}{(1+a)(1+a^2)\cdots(1+a^n)}\ \ \text{for} \ \ a >0 $$

Let: $$ \begin{cases} x_n = \frac{a^n}{(1+a)(1+a^2)\cdots(1+a^n)}\\ n\in\mathbb N \end{cases} $$

By definition of exponent we may rewrite $x_n$ as: $$ x_n = \frac{a}{1+a}\cdot\frac{a}{1+a^2}\cdots\frac{a}{1+a^n} $$

Since $a>0$: $$ \forall a>0,k\in\mathbb N: \frac{a}{1+a^k} < 1 $$

Choose $q_n$ to be in the following form: $$ q_n = \max\left\{\frac{a}{1+a}, \frac{a}{1+a^2}, \dots,\frac{a}{1+a^n}\right\} $$

It follows that: $$ x_n = \frac{a^n}{(1+a)(1+a^2)\cdots(1+a^n)} \le q^n_n $$ Also $x_n > 0$. Now squeezing $x_n$ one may obtain:

$$ 0 \le \lim_{n\to\infty} x_n \le \lim_{n\to\infty} q^n_n = 0 $$ Thus: $$ \lim_{n\to\infty}\frac{a^n}{(1+a)(1+a^2)\cdots(1+a^n)} = 0 $$

Have i missed something?

Answer 1

HINT

As an alternative by ratio test

$$\frac{x_{n+1}}{x_n}=\frac{a^{n+1}}{(1+a)(1+a^2)\cdots(1+a^{n+1})}\frac{(1+a)(1+a^2)\cdots(1+a^n)}{a^n}=\frac{a}{1+a^{n+1}}$$

Answer 2

Following your idea to apply the squeeze theorem, the proof becomes simpler if we note that $0\leq a<1$, $$0\leq \frac{a^n}{(1+a)(1+a^2)\cdots(1+a^n)}\leq a^n\to 0$$ and for $a\geq 1$, $$0\leq \frac{a^n}{(1+a)(1+a^2)\cdots(1+a^n)}\leq \frac{a^n}{2^n a^{n(n+1)/4}}\to 0$$ where we used the fact that $1+a^k\geq 2a^{k/2}.$