Exercise 1.
Let $\alpha\in[0,1]$ and define
\[
a_1:=\frac{\alpha}{2},\quad a_{n+1}:=\frac{a_n^2+\alpha}{2}.
\]Investigate the convergence of $\{a_n\}_{n=1,\infty}$.
(a) $0\leq a_n\leq 1-\sqrt{1-\alpha}$.
By induction.
(i) $n=1$ $\checkmark$
$\quad a_1=\frac{\alpha}{2}\leq 1-\sqrt{1-\alpha}$ $\Leftrightarrow$ $\sqrt{1-\alpha}\leq\frac{2-\alpha}{2}$ $\Leftrightarrow$ $4(1-\alpha)\leq(2-\alpha)^2=4-4\alpha+\alpha^2$ $\Leftrightarrow$ $0\leq\alpha^2$.
(ii) $n$ $\checkmark$
(iii) $n+1$ $\checkmark$
$a_{n+1}=\frac{a_n^2+\alpha}{2}\leq\frac{(1-\sqrt{1-\alpha})^2+\alpha}{2}=1-\sqrt{1-\alpha}$
(b) $a_n\leq a_{n+1}$.
Indeed,
$a_n\leq a_{n+1}$ $\Leftrightarrow$ $a_n\leq\frac{a_n^2+\alpha}{2}$ $\Leftrightarrow$ $0\leq a_n^2-2a_n+\alpha$ $\Leftrightarrow$ $1-\alpha\leq (a_n-1)^2=(1-a_n)^2$ $\Leftrightarrow$ $\sqrt{1-\alpha}\leq 1-a_n$ $\Leftrightarrow$ $a_n\leq 1-\sqrt{1-\alpha}$.
(c) From (a) and (b) it follows that $H:=\lim_{n\to\infty}a_n$ exists and finite. Taking the limit in recursion we obtain
\[
H=\frac{H^2+\alpha}{2}
\] which gives, taking into account (a), that
\[
\lim_{n\to\infty}a_n=1-\sqrt{1-\alpha}.
\]
By induction.
(i) $n=1$ $\checkmark$
$\quad a_1=\frac{\alpha}{2}\leq 1-\sqrt{1-\alpha}$ $\Leftrightarrow$ $\sqrt{1-\alpha}\leq\frac{2-\alpha}{2}$ $\Leftrightarrow$ $4(1-\alpha)\leq(2-\alpha)^2=4-4\alpha+\alpha^2$ $\Leftrightarrow$ $0\leq\alpha^2$.
(ii) $n$ $\checkmark$
(iii) $n+1$ $\checkmark$
$a_{n+1}=\frac{a_n^2+\alpha}{2}\leq\frac{(1-\sqrt{1-\alpha})^2+\alpha}{2}=1-\sqrt{1-\alpha}$
(b) $a_n\leq a_{n+1}$.
Indeed,
$a_n\leq a_{n+1}$ $\Leftrightarrow$ $a_n\leq\frac{a_n^2+\alpha}{2}$ $\Leftrightarrow$ $0\leq a_n^2-2a_n+\alpha$ $\Leftrightarrow$ $1-\alpha\leq (a_n-1)^2=(1-a_n)^2$ $\Leftrightarrow$ $\sqrt{1-\alpha}\leq 1-a_n$ $\Leftrightarrow$ $a_n\leq 1-\sqrt{1-\alpha}$.
(c) From (a) and (b) it follows that $H:=\lim_{n\to\infty}a_n$ exists and finite. Taking the limit in recursion we obtain
\[
H=\frac{H^2+\alpha}{2}
\] which gives, taking into account (a), that
\[
\lim_{n\to\infty}a_n=1-\sqrt{1-\alpha}.
\]
Exercise 2.
Let $\alpha>0$ and define\[
a_1:=\sqrt{\alpha},\quad a_{n+1}:=\sqrt{\alpha+a_n}. \]
Investigate the convergence of $\{a_n\}_{n=1,\infty}$.
(a) $0\leq a_n\leq \frac{1}{2}+\frac{1}{2}\sqrt{1+4\alpha}$.
By induction.
(i) $n=1$ $\checkmark$
$\quad a_1=\sqrt{\alpha}\leq \frac{1}{2}+\frac{1}{2}\sqrt{1+4\alpha}$ $\Leftrightarrow$ $2\sqrt{\alpha}-1\leq\sqrt{1+4\alpha}$.
If $\alpha\leq\frac{1}{4}$ then the last inequality is satisfied. If $\alpha>\frac{1}{4}$ then we can write
$\Leftrightarrow$ $4\alpha+1-4\sqrt{\alpha}\leq 1+4\alpha$.
(ii) $n$ $\checkmark$
(iii) $n+1$ $\checkmark$
$a_{n+1}=\sqrt{\alpha+a_n}\leq\sqrt{\alpha+\frac{1}{2}+\frac{1}{2}\sqrt{1+4\alpha}}=\frac{1}{2}+\frac{1}{2}\sqrt{1+4\alpha}$
(b) $a_n\leq a_{n+1}$ $\Leftrightarrow$ $a_n\leq\sqrt{\alpha+a_n}$ $\Leftrightarrow$ $a_n^2\leq\alpha+a_n$ $\Leftrightarrow$ $a_n^2-a_n+\frac{1}{4}\leq\alpha+\frac{1}{4}$ $\Leftrightarrow$ $\left(a_n-\frac{1}{2}\right)^2\leq\alpha+\frac{1}{4}$.
If $0\leq a_n\leq\frac{1}{2}$ then the last inequality is satisfied. If $a_n>\frac{1}{2}$ then we get
$\Leftrightarrow$ $a_n\leq \frac{1}{2}+\frac{1}{2}\sqrt{1+4\alpha}$ which was proved in (a).
(c) From (a) and (b) it follows that $H:=\lim_{n\to\infty}a_n$ exists and finite. Taking the limit in recursion we obtain
\[
H=\sqrt{\alpha+H}
\]
which gives, taking into account (a), that
\[
\lim_{n\to\infty}a_n=\frac{1}{2}+\frac{1}{2}\sqrt{1+4\alpha} .
\]
By induction.
(i) $n=1$ $\checkmark$
$\quad a_1=\sqrt{\alpha}\leq \frac{1}{2}+\frac{1}{2}\sqrt{1+4\alpha}$ $\Leftrightarrow$ $2\sqrt{\alpha}-1\leq\sqrt{1+4\alpha}$.
If $\alpha\leq\frac{1}{4}$ then the last inequality is satisfied. If $\alpha>\frac{1}{4}$ then we can write
$\Leftrightarrow$ $4\alpha+1-4\sqrt{\alpha}\leq 1+4\alpha$.
(ii) $n$ $\checkmark$
(iii) $n+1$ $\checkmark$
$a_{n+1}=\sqrt{\alpha+a_n}\leq\sqrt{\alpha+\frac{1}{2}+\frac{1}{2}\sqrt{1+4\alpha}}=\frac{1}{2}+\frac{1}{2}\sqrt{1+4\alpha}$
(b) $a_n\leq a_{n+1}$ $\Leftrightarrow$ $a_n\leq\sqrt{\alpha+a_n}$ $\Leftrightarrow$ $a_n^2\leq\alpha+a_n$ $\Leftrightarrow$ $a_n^2-a_n+\frac{1}{4}\leq\alpha+\frac{1}{4}$ $\Leftrightarrow$ $\left(a_n-\frac{1}{2}\right)^2\leq\alpha+\frac{1}{4}$.
If $0\leq a_n\leq\frac{1}{2}$ then the last inequality is satisfied. If $a_n>\frac{1}{2}$ then we get
$\Leftrightarrow$ $a_n\leq \frac{1}{2}+\frac{1}{2}\sqrt{1+4\alpha}$ which was proved in (a).
(c) From (a) and (b) it follows that $H:=\lim_{n\to\infty}a_n$ exists and finite. Taking the limit in recursion we obtain
\[
H=\sqrt{\alpha+H}
\]
which gives, taking into account (a), that
\[
\lim_{n\to\infty}a_n=\frac{1}{2}+\frac{1}{2}\sqrt{1+4\alpha} .
\]
