Prove that following recursively defined sequence converges and its limit is 1/2 
\begin{cases}a_1 = 1\\ \\ \displaystyle a_{1+n} = \sqrt{\sum_{i=1}^n a_i}\end{cases}
Prove that $\{\frac{a_n}{n}\}$ is convergent and its limit is $\frac12$

My proof:   We can recursively define relation between $a_{1+n}$ & $a_n$ as
$a_{1+n} = \sqrt{a_n^2+a_n}$
Also, it is evident that $a_{1+n} > a_n > 1$
Thus, $a_{n}^2 - a_{n-1}^2 + a_{n} - a_{n-1} = a_{n} > 1 > 1/4$
Thus,  $a_{n+1} = \sqrt{a_{n}^2  + a_{n}} > \sqrt{a_{n-1}^2+a_{n-1}+1/4} = a_{n-1} + 1/2  $
Similarly, we can show $a_{n+1} = \sqrt{a_{n}^2  + a_n} < a_n + 1/2$
So, we have  $a_1 + \frac{n-1}2 < a_{n+1} < a_1 + \frac{n+1}2$
Using Sandwich Theorem we got limit of $\frac{a_n}n$ as $\frac12$
Is my proof correct.
I will be grateful if you can provide alternative proofs

Try showing via induction that $(a_n)_{n \in \mathbb{N}}$ is monotone and bounded and the solve the fixed point equation (aka $\lim\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} a_{n+1}$).
