The logistic sequence decays like 1/n

Step 1: convergence to $0$. If $u_n\in(0,1)$, then $u_{n+1}=u_n(1-u_n)\in(0,1)$, so all terms lie in $(0,1)$. Also $u_{n+1}-u_n=-u_n^2<0$: the sequence is decreasing and bounded below, hence converges to some $L\ge0$ with $L=L(1-L)$, so $L=0$.

Step 2: a clever quantity.
$$\frac1{u_{n+1}}-\frac1{u_n}=\frac{u_n-u_{n+1}}{u_nu_{n+1}}=\frac{u_n^2}{u_n\cdot u_n(1-u_n)}=\frac1{1-u_n}\longrightarrow1.$$

Step 3: Cesàro. By the Cesàro lemma (part 2), since the differences of $w_n=\frac1{u_n}$ tend to $1$, $\frac{w_n}n\to1$, i.e. $\frac1{nu_n}\to1$.

Conclusion. $nu_n\to1$, i.e. $u_n\sim\frac1n$.

Numerics. For $u_0=\frac12$: $nu_n=0.69$ at $n=10$, $0.94$ at $n=100$, $0.991$ at $n=1000$, $0.9999$ at $n=10^5$.

Remark. The convergence to $0$ is slow (like $\frac1n$) because $f(x)=x(1-x)$ has $f'(0)=1$: this is the borderline case of the speed-of-convergence classification.