Superlinear convergence proof

Question

We say that $p_n$ converges to $p$ $\underline{\it superlinearly}$ if $$ \lim_{n \to \infty} \frac{|p_{n+1}-p|}{|p_n - p|} = 0. $$

Show that if $p_n$ converges superlinearly that $\lim_{n \to \infty} \frac{|p_{n+1}-p_n|}{|p_n - p|} = 1$.

We can rewrite the problem as $$ \lim_{n \to \infty} \frac{|p_{n+1}-p_n|}{|p_n - p|} + \lim_{n \to \infty} \frac{|p_{n+1}-p|}{|p_n - p|}$$ (since, by assumption, the second limit is $0$). Then, combining we get $$ \lim_{n \to \infty} \frac{|p_{n+1}-p_n|+|p_{n+1}-p|}{|p_n - p|}. $$ Since $p > p_{n+1} > p_n$, this is equal to $$ \lim_{n \to \infty} \frac{|p_n-p|}{|p_n - p|} = 1.$$

I am not sure if this is completely correct -- any thoughts please?

Answer 1

This is the right idea but you have a couple errors. First, there is no reason that $p > p_{n+1} > p_n$ should be true. Second, you must be careful splitting up limits like this when you do not yet know that the limits exist. The statement $$\lim a_n+\lim b_n=\lim(a_n+b_n)$$ is only valid if you know that both limits on the left exist.

The second issue is easy to remedy by reorganizing your proof a little, to start from the limits you know to exist and combine them to get the limit you want. The first issue is more serious. You'll need to analyze the different possibilities for how $|p_{n+1}-p_n|$ could be related to $|p_{n+1}-p|$ and $|p_n-p|$. I encourage you to try this on your own before reading the correct proof hidden below.

For all $n$, we have $$|p_{n+1}-p_n|\leq |p_n-p|+|p_{n+1}-p|$$ and also $$|p_{n+1}-p_n|\geq |p_n-p|-|p_{n+1}-p|$$ by the triangle inequality. Thus $$\frac{|p_n-p|}{|p_n - p|} - \frac{|p_{n+1}-p|}{|p_n - p|}\leq \frac{|p_{n+1}-p_n|}{|p_n - p|}\leq \frac{|p_n-p|}{|p_n - p|} + \frac{|p_{n+1}-p|}{|p_n - p|}$$ for all $n$ such that $p_n\neq p$ (which is all sufficiently large $n$ since $\lim_{n \to \infty} \frac{|p_{n+1}-p|}{|p_n - p|}$ exists). The limit of the left and right sides of this inequality are each $1$, so the limit of the middle is $1$ as well by the squeeze theorem.