Convergence of $u_{n+1} =\sqrt{\sum_{k=0}^{n} u_k}$ I am trying to study the convergence of the sequence defined by:
$$u_0 = a>0, u_{n+1} = \sqrt{\sum_{k = 0} ^{n} u_k}$$
I have shown that $$u_{n+1} - {u_n} = \frac{u_n}{u_{n+1}+{u_n}} \tag{$*$}$$ since
$$u_{n+1} - {u_n} = \sqrt{\sum_{k=0}^{n}u_k} - \sqrt{\sum_{k=0}^{n-1}u_k}$$
then rationalize to get to $(*)$.
And since it can be easily shown that (induction) that $u_n>0$ for all $n$ we have that $$u_{n+1}-u_n>0 \Rightarrow u_{n+1}>u_n$$
S0 $(u_n)$ is an increasing sequence. Suppose $(u_n)$ was bounded above, then $u_n \rightarrow u$ for some $u\in\mathbb{R}$. Passing to the limit in $(*)$ we have the $u = 0$, but this is impossible since $u_0>0$ and $(u_n)$ increasing so then $(u_n)$ does not converge.
Is this argument correct?
 A: Another way of seeing this is to note that $(u_n)$ also verifies the recurrence formula
$$u_{n+1}=\sqrt{u_n+u_n^2}$$
and study the function $f:x\mapsto \sqrt{x+x^2}$. This function stabilizes the interval $[0,+\infty[$, is strictly increasing and it is easy to prove that $x>0$ implies $f(x)>x$.
So the sequence $(u_n)$ is strictly increasing, and cannot be bounded because, as you said, the only fixed point by $f$ is $0$.
It would be funny then to find an equivalent of $u_n$ :-)
This is what I found : from $u_{n+1}=\sqrt{u_n+u_n^2}$, you can first derive, as you pointed,
$$u_{n+1}-u_n=\frac{u_n}{u_n+u_{n+1}}< \frac{1}{2}$$
because $(u_n)$ is strictly increasing. This leads to
$$u_n=u_1+\sum_{k=1}^{n-1} u_{k+1}-u_k\le 1+\frac{n-1}{2}=\frac{n+1}{2}$$
But you can also derive
$$u_{n+1}=u_n\sqrt{1+\frac{1}{u_n}}$$
and because we proved $\lim u_n=+\infty$, we can use the development :
$$u_{n+1}=u_n(1+\frac{1}{2u_n}+o(\frac{1}{u_n})) = u_n+\frac12 +o(1)$$
so 
$$u_{n+1}-u_n=\frac{1}{2}+o(1)\sim \frac12$$ 
and by Cesaro theorem (or by summations of equivalents) :
$$u_n = u_1+\sum_{k=1}^{n-1} u_{k+1}-u_k \sim \frac n2$$
So now you have the limit, an upper bound and an equivalent :-)
A: Since every $u_n$ is positive, we have $u_n \ge \sqrt {u_0}$ for $n \ge 1$, and then $u_n \ge \sqrt {(n-1)\sqrt {u_0}}$ for $n \ge 2$, which shows that $(u_n)$ diverges.
A: let
$$v_n=\frac{\sum_{k=0}^n u_k}{n+1}$$.
we have
$$\frac{u_{n+1}^2}{n+1}=v_n$$.
if $lim_{n\to +\infty}u_n=L$ then
$0=\lim_{n\to+\infty} v_n=L$
using Cesaro average.
Now, if $(u_n)$ is increasing and
$u_0=a>0$, the limit can't be $0$.
thus $(u_n)$ diverges.
OR WITHOUT CESARO
we have
$$u_{n+1}-u_n=\frac{u_n}{u_n+u_{n+1}}$$
and  when $n\to +\infty$
$$L-L=\frac{L}{2L}$$
which proves that 
$(u_n)$ diverges.
