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

As pointed out in question, the sequence satisfy the recurrence relation $a_{n+1} = \sqrt{a_n^2 + a_n}$.
From this, it is easy to see $a_n$ is strictly increasing and positive.
One consequence of this is $a_n$ cannot be bounded.
Assume the contrary, then $a_n$ converges to some finite value $M > 0$. Since the map $x \mapsto \sqrt{x(x+1)}$ is continuous, the "limit" $M$ will then satisfy $M = \sqrt{M^2 + M} > M$ which is absurd.
Another consequence is 
$$a_n < a_{n+1} < \sqrt{a_n^2 + a_n + \frac14} = a_n + \frac12
\quad\implies\quad 1 < \frac{a_{n+1}}{a_n} < 1 + \frac1{2a_n}$$
Since $a_n$ is unbounded, this leads to
$\displaystyle\;\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = 1$ and hence
$$\lim_{n\to\infty} \frac{a_{n+1}-a_{n}}{(n+1)-n} = 
\lim_{n\to\infty}\frac{a_{n+1}^2-a_n^2}{a_{n+1}+a_n}
= \lim_{n\to\infty}\frac{a_n}{a_n + a_{n+1}}
= \lim_{n\to\infty}\frac{1}{1 + \frac{a_{n+1}}{a_n}}
= \frac{1}{1+1} = \frac12$$
By Stolz-Cesàro, we can conclude
$$\lim_{n\to\infty} \frac{a_n}{n} \quad\text{exists and equals to}\quad \lim_{n\to\infty}\frac{a_{n+1}-a_n}{(n+1)-n} = \frac12$$
