# Let $a_{n+1}=\sqrt{a_1+a_2+\cdots+a_n}$ .Prove that $\lim\limits_{n \rightarrow \infty} \frac{a_n}{n}=\frac{1}{2}$

A sequence $a_n$ is defined as $a_1=1$ and $a_{n+1}=\sqrt{a_1+a_2+\cdots+a_n}$ .Prove that $\lim\limits_{n \rightarrow \infty} \frac{a_n}{n}=\frac{1}{2}$

I have no idea how to approach this. But I have a feeling that Cesaro's lemma may come in handy

• Using the recursion, you can get $a_{n+1}^2 = a_n^2+a_n$. This may or may not work, but you might be able to prove by induction that $n/2-f(n) \le a_n \le n/2+g(n)$ for some $f(n),g(n)$ which are $o(n)$. – JimmyK4542 Dec 23 '16 at 18:28
• I would try lots of different approaches to this. One approach is to use the triangle inequality, then see if the result of $a_{n+1}$ having an upper bound helps. Eg is the upper bound / n convergent? If yes by the comparison test this implies $a_{n+1}/ n$ is convergent. To get the limit you will have to think and use one or more various theorems and lemmas. L'hopitals rule may be worth a try also. – unseen_rider Dec 23 '16 at 18:38

Easy to show $a_{n+1}^2 = a_n^2+a_n$ as @JimmyK4542 pointed out.
Because $a_n$ is increasing, it has a limit. Suppose $a_n$ is convergent to $L$. Then $L^2=L^2 + L$ therefore $L=0$ absurd. It follows $a_n$ is divergent and $\lim a_n = +\infty$.
From $(\frac {a_{n+1}} {a_n})^2 = 1 + \frac 1 {a_n}$ it follows $\lim \frac {a_{n+1}} {a_n}=1$
Also from $$a_{n+1} - a_n = \frac {a_n} {a_{n+1} + a_n}$$
we have $$\lim (a_{n+1} - a_n) = \lim \frac {1} {\frac {a_{n+1}} {a_n} + 1} = \frac 1 2$$