Show that if {pn} converges to p with order α>1, then {pn} converges superlinearly to p.
Say
{pn}→p with
order α>1. Then we know that, for some
λ and for large
n,
∣pn−p∣α∣pn+1−p∣≈λhence
∣pn−p∣∣pn+1−p∣≈λ⋅∣pn−p∣α−1
If we take a
limit of both sides, then the left-hand side becomes the
limit in the definition of superlinear
convergence. The right-hand side vanishes to
zero because
{pn} is known to converge to
p and because
α−1>0.
Thus the condition for superlinear
convergence is met!