Existence or non existence of a sequence $0let $u_n$ be a sequence such that $\sum_{n=0}^{\infty}u_n$ is a convergent series and $\sum_{n=0}^{\infty}u_n=s$ , Now I want to know if there exist a sequence  $0<u_n<1$ such that: $\sum_{n=0}^{\infty}\sqrt{u_n}=\sqrt{s}$ if  if $\sum_{n=0}^{\infty}u_n=s$  or to disprove its existence ?
Edit I have edited the question because I do not want nul terms and keep the convergence of the titled series ,Thanks to Eric Towers of him attention in below comment
 A: 
Lemma: Let $x_1,x_2 > 0$.
  Then
$$\sqrt{x_1}+\sqrt{x_2} > \sqrt{x_1 + x_2}$$

Proof: Both sides are positive.
Square them and compare. $\square$

Let $U_n = \sum_{i=0}^n u_n$ and $S_n = \sum_{i=0}^n \sqrt u_n$.
By the lemma, we have that $a = S_1 - \sqrt {U_1} > 0$.
We prove by induction that $S_n - \sqrt{U_n}\geqslant a > 0$, which answers the question negatively.
The base case is of course provided above, so we proceed to the induction step.
Suppose that $S_n - \sqrt{U_n}\geqslant a$.
We will show that this also holds for $n+1$.
We have
$$\begin{align}
S_{n+1} - \sqrt{U_{n+1}}
&=
\underbrace{S_n - \sqrt{U_n}}_{\text{induction hypothesis; }\geqslant a} + \sqrt{U_n} + \sqrt{u_{n+1}} - \sqrt{U_{n+1}}
\\&\geqslant
a + \underbrace{\sqrt{U_n} + \sqrt{u_{n+1}} - \sqrt{U_n + u_{n+1}}}_{\text{lemma; } > 0}
> a > 0,
\end{align}
$$
which completes the induction.

Let $X_n = S_n - \sqrt{U_n}$ for $n\geqslant 1$.
The proof shows that $(X_n)$ is bounded below by $a = X_1$, but upon closer inspection it's easy to see that $(X_n)$ is strictly increasing.
If we relax the condition $u_n > 0$ to $u_n\geqslant 0$, we see that $(X_n)$ is non-decreasing.
It follows that the conclusion still holds in this case provided that $X_n > 0$ for some $n$, that is, provided that $u_n>0$ for at least two indices.
A: If we allow one negative term, I have a solution. Let $V= \frac{1}{12} \left(-2+\sqrt[3]{172-12 \sqrt{177}}+2^{2/3} \sqrt[3]{43+3 \sqrt{177}}\right)\approx 0.602785.$ Let $u_n = ((n-1)V^n))^2$, with $\sqrt{u_0}=-1$. Then using the derivatives of the geometric series, you can show 
$$
\sqrt{\sum_{n=0}^{\infty}u_n} = \sum_{n=0}^{\infty} \sqrt{u_n}
$$I found $V$ by using a parameter $r$, playing around with $u_n =\text{poly}(n) r^n$ and solving for $r$.
A: No. Let $a_n > 0$ and $\sum_{n= 0}^\infty a_n = r$ (i.e. the series $\sum_{n= 0}^\infty a_n$ converges absolutely to $r$). Then the series $\sum_{n= 0}^\infty  c_n $ with $c_n = \sum_{k=0}^n a_ka_{n-k}$ is the Cauchy product of $\sum_{n= 0}^\infty a_n$ with itself and converges to $r^2$. That is, $\sum_{n= 0}^\infty  c_n = r^2$. All $c_ n > 0$, thus $\sum_{n= 0}^\infty  c_n > \sum_{m= 0}^\infty  c_{2m}$. But $c_{2m} > a_m^2$ for $m >0$, thus $r^2 >  \sum_{m= 0}^\infty  a^2_m$.
The situation is the same if we require $a_n \ge 0$ for all $n$ and $a_n > 0$ for at least two indices $n$.
If we allow negative $a_n$, then it is possible that $\sum_{n= 0}^\infty a_n = r$ and $\sum_{n= 0}^\infty a^2_n = r^2$. Let $a_n > 0$ for $n \ge 1$ such that $\sum_{n= 1}^\infty a_n = s_1 < \infty$. Then $\sum_{n= 1}^\infty a^2_n$ is also convergent and $\sum_{n= 1}^\infty a^2_n = s_2 < s^2_1$ (see the above arguments). Define
$$a_0 = \frac{s_2 - s^2_1}{2s_1} .$$
Then $a_0 < 0$ and
$$\sum_{n= 0}^\infty a_n = s_1 + \frac{s_2 - s^2_1}{2s_1} = \frac{s_2 + s^2_1}{2s_1} ,$$
$$\sum_{n= 0}^\infty a^2_n = s_2 + \left( \frac{s_2 - s^2_1}{2s_1} \right)^2 = \frac{s^2_2 + 2s_2s_1^2 +s^4_1}{4s^2_1} =  \left( \frac{s_2 + s^2_1}{2s_1} \right)^2 .$$
A: HINT:
$$\sum \frac{u_n}{s}= 1 $$
and
$$\sum \sqrt{\frac{u_n}{s}} = 1$$
