Finding the limits related to $a_1=1$, $a_{k+1}=\sqrt{a_1+a_2+\cdots +a_k}$ Suppose $a_1=1, a_{k+1}=\sqrt{a_1+a_2+\cdots +a_k}, k \in \mathbb{N}$. Find the limits
$$i)\space \lim_{n\to\infty}\displaystyle \frac{\sum_{k=1}^{n} a_{k}}{n\sqrt{n}}$$
$$ii)\space \lim_{n\to\infty}\displaystyle \frac{\sum_{k=1}^{n} a_{k}}{n^2}$$
I'm puzzled with it. What to do?
 A: For $n\ge1$, we have
$$
\sum_{k=1}^na_k=a_{n+1}^2\tag{1}
$$
$(1)$ leads to $a_{n+1}^2-a_n^2=a_n$ which implies that $a_n$ is increasing and that
$$
a_{n+1}-a_n=\frac{a_n}{a_{n+1}+a_n}\le\frac12\tag{2}
$$
In fact,
$$
\frac12-\frac{a_n}{a_{n+1}+a_n}=\frac12\frac{a_{n+1}-a_n}{a_{n+1}+a_n}\le\frac12\frac{\frac12}{2a_n}=\frac1{8a_n}\tag{3}
$$
Thus, $(2)$ and $(3)$ say that
$$
\frac38\le\frac12-\frac1{8a_n}\le a_{n+1}-a_n\le\frac12\tag{4}
$$
Summing $(4)$ yields
$$
\frac{3n+5}{8}\le a_n\le\frac{n+1}{2}\tag{5}
$$
We can use $(5)$ to improve $(4)$:
$$
\frac12-\frac1{3n+5}\le a_{n+1}-a_n\le\frac12\tag{6}
$$
Since
$$
\sum_{k=1}^{n-1}\frac1{3n+5}\le\int_0^{n-1}\frac{\mathrm{d}x}{3x+5}=\frac13\log\left(\frac{3n+2}{5}\right)\tag{7}
$$
Summing $(6)$ yields
$$
\frac{n+1}{2}-\frac13\log\left(\frac{3n+2}{5}\right)\le a_n\le\frac{n+1}{2}\tag{8}
$$

To handle $i)$,
$$
\begin{align}
\frac1{n^{3/2}}\sum_{k=1}^na_n
&=\frac1{n^{3/2}}a_{n+1}^2\\
&\ge\frac1{n^{3/2}}\left(\frac{n+2}{2}-\frac13\log\left(\frac{3n+5}{5}\right)\right)^2\\[12pt]
&\to\infty\tag{9}
\end{align}
$$
To handle $ii)$,
$$
\begin{align}
\frac1{n^2}\sum_{k=1}^na_n
&=\frac1{n^2}a_{n+1}^2\tag{10}
\end{align}
$$
Applying $(8)$ yields
$$
\frac1{n^2}\left(\frac{n+2}{2}-\frac13\log\left(\frac{3n+5}{5}\right)\right)^2
\le\frac1{n^2}a_{n+1}^2
\le\frac1{n^2}\left(\frac{n+2}{2}\right)^2\tag{11}
$$
So by the Squeeze Theorem, we have
$$
\lim_{n\to\infty}\frac1{n^2}\sum_{k=1}^na_n=\frac14\tag{12}
$$
A: For $n > 1$, we have
$$a_{n+1}^2 = a_{n} + ( a_{n-1} + \ldots + a_{1} ) = a_{n} + a_{n}^2$$
This implies
$$\displaystyle a_{n+1} = \sqrt{a_n^2 + a_n} < \sqrt{a_n^2 + a_n +\frac14} = a_n + \frac12$$
Notice $a_2 = \sqrt{a_1} = 1$, this gives us an upper bound $a_m \le \frac{m}{2}$ for $m > 1$.
As pointed out by robjohn, there is a mistake in original derivation of a lower bound.
To obtain a correct lower bound, we will construct it in 3 stages.


*

*It is clear $a_k$ is an increasing sequence and since $a_1 = 1$, we have 
$a_k \ge 1$ for all $k$.  

*Since $\quad a_{n+1} = \sqrt{ a_n^2 + a_n } \ge \sqrt{ a_n^2 + \frac23 a_n + \frac13} > a_n + \frac13\quad$ and $a_1 = a_2 = 1 $,
we get $a_m > \frac{m}{3}$ for $m > 0$.  

*Observe $\displaystyle\quad(a_n + \frac12 - \frac{1}{8a_n})^2 =
a_n^2 + a_n - \frac{1}{8a_n} ( 1 - \frac{1}{8a_n}) < a_{n+1}^2.$
The lower bound from $2^{nd}$ stage implies 
$$\begin{align}
& a_{n+1} > a_n + \frac12 - \frac{1}{8a_n} > a_n + \frac12 - \frac{3}{8n}\\
\implies & 
a_m \ge \frac{m}{2} - \frac{3}{8} \sum_{k=2}^{m-1}\frac{1}{k}\;\;\text{ for }m > 2
\end{align}$$


Since $\displaystyle \sum_{k=2}^{m-1}\frac{1}{k} \sim \log m + O(1)$ for large $m$, we find $\displaystyle\quad a_m = \frac{m}{2} + O(\log m)\quad$ as $m \to \infty$.
For the first limit, we have the estimate:
$$\frac{\sum_{k=1}^{n}a_k}{n\sqrt{n}} = \frac{a_{n+1}^2}{n\sqrt{n}} = \frac{\left(\frac{n}{2} + O(\log n)\right)^2}{n\sqrt{n}} \sim O(\sqrt{n})$$
This implies the first limit diverges. For the second limit, we have
$$\frac{\sum_{k=1}^{n}a_k}{n^2} = \frac{a_{n+1}^2}{n^2} = \frac{\left(\frac{n}{2} + O(\log n)\right)^2}{n^2} = \frac14 + O(\frac{\log n}{n})$$
Since $\lim_{n\to\infty} \frac{\log n}{n} = 0$, the second limit exists and equal to $\frac14$.
