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

Solution 1:

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 $$\lim (a_{n+1} - a_n) = \lim \frac {1} {\frac {a_{n+1}} {a_n} + 1} = \frac 1 2$$

Now use Cesaro's theorem to conclude.