When ${u_n}$ is converge, find $u_0$ and solve $\lim(nu_n)$ with $ {u_n} : 2u_{n+1}-2u_n+u_n^2=0$ With ${u_n}: 2u_{n+1}-2u_n+u_n^2=0$. its show $${u_{n + 1}} - {u_n} = \dfrac{{ - u_n^2 + 2{u_n}}}{2} - {u_n} =  \dfrac{- u_n^2}{2} \leqslant  0$$
${u_n}$ is decrease.
When $n \to \infty $ we have $\lim(u_n)=0$ (it's happen when ${u_n}$ is converge).

Also have
$${u_{n + 1}} = \frac{{ - u_n^2 + 2{u_n} - 1}}{2} + \frac{1}{2} =  - \frac{{{{\left( {{u_n} - 1} \right)}^2}}}{2} + \frac{1}{2} \leqslant \frac{1}{2}$$
It's true with ${u_{n+1}}$ so $u_n\leqslant\dfrac{1}{2}$. Therefore I guest $u_0=\dfrac{1}{2}$ but i have no idea no prove it  and how to find $\lim(nu_n)$.

If the $\left\{ u_n \right\} $ is converge,let $\displaystyle\lim_{n\rightarrow \infty} u_n=A$,we can get that$$A=-\frac{A^2}{2}-A\Rightarrow A=0 \text{or} -4$$
i.e. $\displaystyle\lim_{n\rightarrow \infty} u_n=0$ or $\displaystyle\lim_{n\rightarrow \infty} u_n=-4$.
Due to $\displaystyle u_{n+1}-u_n=-\frac{u_{n}^{2}}{2}\leqslant 0$
If we want to the $\left\{ u_n \right\} $ is converge,we should let the $\left\{ u_n \right\} $ has the lower bound.
$$u_{n+1}=-\frac{u_{n}^{2}}{2}+u_n=-\frac{\left( u_n-1 \right) ^2-1}{2}=-\frac{\left( u_n-1 \right) ^2}{2}+\frac{1}{2}$$
If $\displaystyle 0<u_0<2$,then $\displaystyle 0<u_1<\frac{1}{2}$,so on and so forth,we can see that $\displaystyle 0<u_n<\frac{1}{2}$.
In this case,the $\left\{ u_n \right\} $ is converge($\displaystyle \lim_{n\rightarrow \infty} u_n=0$),and we come to calculate the $\displaystyle\lim_{n\rightarrow \infty} nu_n$($\text{Stolz}$).
$$
\begin{align*}
\lim_{n\rightarrow \infty} nu_n&=\lim_{n\rightarrow \infty} \frac{n}{\frac{1}{u_n}}
\\
&=\lim_{n\rightarrow \infty} \frac{1}{\frac{1}{u_{n+1}}-\frac{1}{u_n}}
\\
&=\lim_{n\rightarrow \infty} \frac{1}{\frac{2}{2u_n-u_{n}^{2}}-\frac{1}{u_n}}
\\
&=\lim_{n\rightarrow \infty} \frac{1}{\frac{2}{u_n\left( 2-u_n \right)}-\frac{1}{u_n}}
\\
&=\lim_{n\rightarrow \infty} \frac{u_n\left( 2-u_n \right)}{2-2+u_n}
\\
&=\lim_{n\rightarrow \infty} \left( 2-u_n \right) 
\\
&=2
\end{align*}
$$
