Question

If $a_n=\sqrt{1+\sqrt{1+\sqrt{1+\cdots }}}$ having $n$ radical signs, then by method of mathematical induction which of the following is/are true

• A. $a_n<1,\forall n\in N$
• B. $a_n<2,\forall n\in N$
• C. $a_n\ge 1,\forall n\in N$
• D. $a_n>2,\forall n\in N$

Answer 1

Answer: (B), (C)

$a_n\ge 1,\forall n\in N$

Given $a_n=\sqrt{1+\sqrt{1+\sqrt{1+\cdots }}}$ having $n$ radical signs
$a_1=\sqrt{1}=1$;
Let $a_m<2$.
Then $a_{m+1}=\sqrt{1+a_m}$
$\Rightarrow a_{m+1}^2=1+a_m<1+2=3$
$\Rightarrow a_{m+1}<\sqrt{3}<2$
So $a_n<2,\forall n\ge 1$