If $x_{n+1}=f(x_n)$ and $x_{n+1}-x_n\to 0$, then $\{x_n\}$ converges Let $f:[0,1] \rightarrow [0,1]$be a continuous function. Choose any point $x_0 \in [0,1]$ and define a sequence recursively by $x_{n+1}=f(x_n)$. Suppose $\lim_{n \rightarrow \infty}x_{n+1}-x_n =0$, does this sequence converge?

I'm fairly sure it need not converge.
Let $$H_n = \sum_{i=1}^n \frac{1}{i}$$
Define $y_n = H_n \pmod{1}$ if $\lfloor H_n \rfloor$ is even, and $1-H_n \pmod{1}$ otherwise. This sequence oscillates between 0 and 1 indefinitely, so it doesn't converge; although $y_{n+1} - y_n$ tends to $0$. It is the case that the difference between distinct harmonic numbers is never an integer, according to Wikipedia, so this sequence never duplicates any value.
We just need to show that we can define continuous $f: [0,1] \to [0,1]$ such that $f(x_n) = x_{n+1}$.
I've convinced myself of this with some rather dubious reasoning about infinitesimals.
