<s>[INST] 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?
 [/INST] 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
$$
\lim_{n\to\infty}\frac1{n^2}\sum_{k=1}^na_n=\frac14\tag{12}
$$
</s>