I am preparing for a test and wanted to ask you
$a_0 = 1; a_{n+1} = \sqrt{a_n} + \frac{15}{4} $
I already showed its strictly monotonically increasing. Now im trying to calculate the limit.
$$\lim a_{n+1} = \lim a_n \Leftrightarrow a = \sqrt{a} + \frac{15}{4} \Leftrightarrow a = (a- \frac{15}{4})^2 \Leftrightarrow 0 = a^2 - \frac{17a}{2} + \frac{225}{16}$$
$$\Longrightarrow a_1 = 2.25 , a_2 = 6.25$$ So you basically take the first limit $a_1 = 2.25$ . Is that correct? Is there better way of calculating the limit? Thank you
The missing part is $a_n$ has upper bound.
Using induction, we'll show $a_n \le \frac {25} 4, \forall n \ge 1$.
For $n=1$ it's obvious. Suppose it's true for $n$. Then $a_{n+1}=\sqrt{a_n} + \frac{15}{4} \le \frac 5 2 + \frac {15} 4 = \frac {25} 4$