Answer
(a) The general term is ${a_n} = {\left( {0.5} \right)^{{2^n}}}$, for $n = 0,1,2,...$.
(b) The results suggest that the limit of the sequence may be $0$.
(c) $\mathop {\lim }\limits_{n \to \infty } {\left( {\dfrac{1}{2}} \right)^{{2^n}}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{{2^{{2^n}}}}} = 0$
This confirm our conjecture in part (b) is correct.
(d) The sequence converges if $0 \lt {a_0} \lt 1$.
Work Step by Step
(a)
${a_0} = 0.5$
${a_1} = {\left( {0.5} \right)^2}$
${a_2} = {\left( {{{\left( {0.5} \right)}^2}} \right)^2} = {\left( {0.5} \right)^{{2^2}}}$
${a_3} = {\left( {{{\left( {{{\left( {0.5} \right)}^2}} \right)}^2}} \right)^2} = {\left( {0.5} \right)^{{2^3}}}$
$...$
So, the general term is ${a_n} = {\left( {0.5} \right)^{{2^n}}}$, for $n = 0,1,2,...$.
(b) Using a calculating utility, we compute ${\left( {0.5} \right)^{{2^n}}}$ and list them in the table below:
$\begin{array}{*{20}{c}}
n&{{{\left( {0.5} \right)}^{{2^n}}}}\\
0&{0.5}\\
1&{0.25}\\
2&{0.0625}\\
3&{0.00390625}\\
4&{0.0000152588}\\
5&{2.32831 \times {{10}^{ - 10}}}
\end{array}$
The results suggest that the limit of the sequence may be $0$.
(c) To confirm our conjecture in part (b), we evaluate the limit of ${a_n} = {\left( {0.5} \right)^{{2^n}}}$ as $n \to \infty $.
$\mathop {\lim }\limits_{n \to \infty } {\left( {\dfrac{1}{2}} \right)^{{2^n}}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{{{2^{{2^n}}}}} = 0$
This confirm our conjecture in part (b) is correct.
(d) Let ${a_0} = b$. Thus, the formula for the general term is ${a_n} = {b^{{2^n}}}$, for $n=0,1,2,...$.
Let $x = {2^n}$. If $n \to \infty $, then $x \to \infty $.
Recall that if $b \gt 1$, then ${b^x} \to \infty $ as $x \to \infty $, and if $0 \lt b \lt 1$, then ${b^x} \to 0$ as $x \to \infty $. Therefore, this procedure produce a finite limit which is zero if $0 \lt b \lt 1$. Thus, the sequence converges if $0 \lt {a_0} \lt 1$.
