Recursive sequence of square roots of previous elements

Question

Bruckner, Bruckner, Thompson - Elementary Real Analysis

$a_1 = 1$ and $a_{n+1} = \sqrt{a_1+ a_2 + .. + a_n}$

Show that $$\lim_{n \to \infty}\ \frac{a_n}{n} = \frac12$$

I cannot untangle the square root of the sum, to show that it 'converges' to $\frac{n}2$. Any help much appreciated.

Answer 1

We may prove for first that: $$ a_{n+1}^2 = a_{n}^2 + a_{n}\leq \left(a_n+\frac{1}{2}\right)^2 \tag{1}$$ from which it follows that $\{a_n\}_{n\geq 1}$ is increasing and $a_n\leq \frac{n+1}{2}$. $(1)$ also gives: $$ a_{n+1}-a_n = \frac{a_n}{a_n+a_{n+1}}=\frac{1}{2}\left(1-\frac{a_{n+1}-a_{n}}{a_n+a_{n+1}}\right)\tag{2} $$ but since $a_{n+1}-a_n\leq\frac{1}{2}$, the previous line gives: $$ a_{n+1}-a_n \geq \frac{1}{2}-\frac{1}{4(a_n+a_{n+1})}\geq\frac{1}{2}-\frac{1}{8a_n} \tag{3}$$ and we may check by induction that: $$ a_n \geq \frac{n+1}{2}-\frac{\log(n+1)}{4}.\tag{4}$$ By squeezing, it follows that: $$ \lim_{n\to +\infty}\frac{a_n}{n}=\color{red}{\frac{1}{2}}.\tag{5}$$

Answer 2

Hint use stolz $$I=\lim_{n\to\infty}\dfrac{a_{n}}{n}=\lim_{n\to\infty}(a_{n+1}-a_{n})$$ and Note $$a_{n}=S_{n}-S_{n-1}=a^2_{n+1}-a^2_{n}$$

so $$a_{n+1}=\sqrt{a^2_{n}+a_{n}}$$ it is not easy to prove $a_{n}\to+\infty,n\to\infty$ so $$I=\lim_{n\to\infty}(\sqrt{a^2_{n}+a_{n}}-a_{n})=\lim_{x\to+\infty}(\sqrt{x^2+x}-x)=\dfrac{1}{2}$$