# Understanding the proof: If $a_1=1$ and $a_{n+1}=\sqrt{a_1+a_2+\cdots+a_n}.$ Then, $\lim\limits_{n\to\infty}\frac{a_n}{n}=\frac{1}{2}$

I want to understand a section in the following proof, let $a_1=1$ and $$a_{n+1}=\sqrt{a_1+a_2+\cdots+a_n},$$ then, $$\lim\limits_{n\to\infty}\frac{a_n}{n}=\frac{1}{2}.$$

Here is my question: the above question was solved in Let $a_{n+1}=\sqrt{a_1+a_2+\cdots+a_n}$ .Prove that $\lim\limits_{n \rightarrow \infty} \frac{a_n}{n}=\frac{1}{2}$, by taking that $a_{n+1}^2=a_{n}^2+a_n.$ Please, can you show me how $a_{n+1}^2=a_{n}^2+a_n$ was gotten?

• The linked question is different. It deals with a square root; your's has the nth root. – user58697 Jul 15 '18 at 23:43
• Well, you can begin with $a_{n+1}^2 = a_1 + \cdots + a_n$ and $a_n^2 = a_1 + \cdots + a_{n-1}$. – Sangchul Lee Jul 15 '18 at 23:45
• @ user58697 and spaceisdarkgreen: Sorry it was a typo! I've corrected it! – Omojola Micheal Jul 15 '18 at 23:45
• @Mike they don't assume it. They prove it – mathworker21 Jul 15 '18 at 23:50
• @mathworker21: Sorry for the language misuse! – Omojola Micheal Jul 16 '18 at 0:52

Note that $$a_{n+1}^2 = a_1+\ldots +a_{n-1} + a_n= \left(\sqrt{a_1+\ldots +a_{n-1}}\right)^2 + a_n= a_n^2+a_n$$