Existence or non existence of a sequence $0<u_n<1$ such that $\sum_{n=0}^{\infty}\sqrt{u_n}=\sqrt{s}$ if $\sum_{n=0}^{\infty}u_n=s$?

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. $\square$

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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.

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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.