How to evaluate $\lim_{n\to\infty} a_n$, where $a_{n+1} = \sqrt{1+\frac12 a_n}$? Note: A similar question (same recursive function) has been asked here, but none of the answers is relevant to my question.

I am trying to evaluate $\lim_{n\to\infty} a_n$. The sequence $a_n$ is given by the recursive  function $$a_{n+1} = \sqrt{1+\frac12 a_n}$$ with $$a_1 =0$$


*

*I proved using induction that the monotonicity is: $a_n \nearrow$

*and that the upper bound is $a_n> \sqrt{2}$

Theorem 1

If $a_n$ is monotonous and bounded then it converges, thus its limit exists

Theorem 2

If $\lim_{n\to\infty} a_n = M$ then every subsequence has the same limit.


Therefore applying limit to $(1)$:
$$ \lim_{n\to\infty} a_{n+1} = \sqrt{1+\frac12 a_n} \iff M = \sqrt{1+\frac12 \lim_{n\to\infty} a_n} \iff M^2 = 1 +\frac{M}{2} \iff $$
$$ \boxed{\lim_{n\to\infty} a_{n} = M=\frac{1+\sqrt{17}}{4}} $$

This value of the limit is smaller than the upper bound $\sqrt{2}$ and this concerns me.
Does this fact mean that the upper bound I've found is some upper bound  but not the supremum or did I make a mistake in calculating the limit?

Edit: Upper bound proof
We will prove by induction that $a_n < \sqrt{2}$.

*

*For $n=1$: $a_2 = \sqrt{1 +\frac12 1} < \sqrt{2}$

*For $n=k$: Let $a_k < \sqrt{2}$

*For $n=k+1:$ $a_{k+1} = \sqrt{1 + \frac12 a_k} < \sqrt{1 + \frac{\sqrt{2}}{2}} = \sqrt{1 + \frac1{\sqrt{2}}}< \sqrt{2}$
Hence indeed $a_n < \sqrt{2}$
 A: Yes, it's just an upper bound, not a supremum. Indeed, you can prove that the sequence is no terms exceed $\frac{3}{2}$ by induction. We have $a_2 = \sqrt{1+ \frac{1}{2}}<\sqrt{1+ \frac{5}{4}}=\frac{3}{2} $. Assume that $a_n < \frac{3}{2}$, it still holds for $a_{n+1}$:
$$a_{n+1} = \sqrt{1+\frac{a_n}{2}} < \sqrt{1+\frac{3}{4}} < \sqrt{1+\frac{5}{4}}=\frac{3}{2}$$
Moreover, in various Calculus 1 textbooks, you can easily find that if $\{a\}_{n\geq 1}$ is a sequence that $\lim_{n\to \infty} a_n$ exists, the sumpremum of $\{a\}_{n\geq 1}$ is the limit of itself.
A: Let $L = \frac{1+\sqrt{17}}{4}$. Then $L^2 = 1 + \frac{1}{2}L$ and $\sqrt{1 + \frac{1}{2}L} = L$.
Let us use the mathematical induction to prove: $0 < a_n < a_{n+1} < L$ for all $n \ge 2$.
For $n = 2$, it is easy to verify it.
Assume that the inequality is true for $n = k$ ($k\ge 2$), i.e., $0 < a_k < a_{k+1} < L$.
Let us prove that the inequality is also true for $n = k + 1$, i.e., $0 < a_{k+1} < a_{k+2} < L$.
First, since $a_{k+1} < L$, we have $a_{k+2} = \sqrt{1 + \frac{1}{2}a_{k+1}} < \sqrt{1 + \frac{1}{2}L} = L$.
Second, since $0 < a_{k+1} < L$ and $L > \frac{1}{2}$, we have $(a_{k+1} - L)(a_{k+1} + L - \frac{1}{2}) < 0$
that is $a_{k+1}^2 - \frac{1}{2}a_{k+1} - 1 < 0$ (using $L^2 = 1 + \frac{1}{2}L$), which results in
$a_{k+2} = \sqrt{1 + \frac{1}{2}a_{k+1}} > a_{k+1}$. This completes the proof.
$\phantom{2}$
Thus, $\lim_{n\to \infty} a_n$ exists (finite). Let $\lim_{n\to \infty} a_n = M > 0$.
Taking limit on both sides of $a_{n+1} = \sqrt{1 + \frac{1}{2}a_n}$, we have
$M = \sqrt{1 + \frac{1}{2}M}$ which results in $M = \frac{1+\sqrt{17}}{4}$. We are done.
