Prove that sequence is divergent Define the sequence $u_{n}$ as follows :
$u_{0}=a>0$,     $\forall n \in \mathbb{N}$ , 
$u_{n+1}=\displaystyle\sqrt{\sum_{k=0}^{n}u_{k}}.$
Prove that the sequence $u_{n}$ diverges.
 A: It is easy to see that all terms of this sequence are non-negative.
Notice that, for $n\ge 1$,
$u_{n+1}=\sqrt{\sum\limits_{k=0}^{n}u_{k}}=\sqrt{u_n+\sum\limits_{k=0}^{n-1}u_{k}}=\sqrt{u_n+u_n^2}$.
This implies
$$u_{n+1}^2=u_n+u_n^2.$$
Thus the sequence $(u_n^2)_{n=1}^\infty$ is increasing  and, consequently, so is $(u_n)_{n=1}^\infty$.
Moreover, we have $u_{n}\ge\sqrt{a}>0$ for $n\ge 1$, which implies that the sequence $(u_n^2)$ is unbounded. 
(Since we have $u_{n+1}^2-u_n^2\ge\sqrt a$ and this implies $u_{n+1}^2 \ge u_1^2 + n\sqrt{a}$.) 
Consequently, $(u_n)$ is unbounded.
A: Alternatively, you could also use that the root function is a strictly monotonous function and that obviously all $u_{n,n\in\mathbb{N}}>0$ due to the monotony of the root function and the fact that $u_{0}>0$. Then, one has to define an upper bound, because even if the sequence is increasing, it is not guaranteed that there doesn't exist an upper bound.
Therefore you can use the following estimate:
$u_{n}>u_{0}=a$ hence $u_{n+1}=\sqrt{\sum_{i=0}^{n}u_{i}} >= \sqrt{\sum_{i=0}^{n}u_{0}} = \sqrt{n\cdot a} = C\cdot\sqrt{n}$, where $C=\sqrt{a}>0$. But it is known that $\sqrt{n}$ is a divergent sequence. As the sequence in the exercise is bounded from below by a divergent sequence, it cannot possess a finite upper bound. Hence it must diverge. Q.E.D.
A: This is essentially Martin's answer again:
A sequence $\{a_n\}$ of positive terms converges if and only if the sequence $\{a_n^2\}$ converges.  
We will show that $\{u_n^2\}$ diverges.
Here,  the sequence $\{u_n\} $ is increasing for $n\ge1$ (since $\{u_n^2\}$ is increasing) and consists of positive terms. Thus,
$$
u_{n+1}^2\ =\  u_0+\sum_{k=1}^n\, u_k \
\ge\ u_0+\sum_{k=1}^n \, u_1 \
=\ a +n\cdot \sqrt a \ \ \buildrel{n\rightarrow\infty}\over\longrightarrow\ \ \infty.  
$$
