Speed of convergence

1. If $a>1$, choose $\rho\in(1,a)$. For large $n$, $v_n>\rho$, so $|u_{n+1}-\ell|>\rho|u_n-\ell|$, and $|u_n-\ell|$ grows like $\rho^n$, contradicting $u_n\to\ell$. So $a\le1$.

2.
- Slow: $u_n=\frac1n$, $\ell=0$, $v_n=\frac n{n+1}\to1$.
- Geometric: $u_n=q^n$ with $0<q<1$, $v_n=q$.
- Fast: $u_n=2^{-2^n}$, $v_n=2^{-2^n}\to0$ (since $u_{n+1}=u_n^2$).

3. By the mean value theorem there is $\xi_n$ between $u_n$ and $\ell$ with
$$u_{n+1}-\ell=f(u_n)-f(\ell)=f'(\xi_n)(u_n-\ell).$$
Since $\xi_n\to\ell$ and $f'$ is continuous, $v_n=|f'(\xi_n)|\to|f'(\ell)|$.

Interpretation. $f'(\ell)=0$ gives fast (superlinear) convergence, as in Newton's method; $0<|f'(\ell)|<1$ gives geometric convergence; $|f'(\ell)|=1$ gives slow convergence (for instance $u_{n+1}=\sin u_n$ tends to $0$ like $\sqrt{3/n}$).