A basic sequence convergence question Consider the following iterative algorithm 
$$x_{n+1} = x_n + a(n)h(x_n)$$
where $x_n$'s are bounded and $h$ is lipschitz( and hence also bounded) and $a_n \rightarrow 0$. Is $x_n$ converges ? Assume some initial condition.
Now consider the function $X^0[0,\infty)$ defined by $X^0(t) = x_n + h(x_n)(t-t_n)$ when $t \in [t_n, t_{n+1}]$, so it is just the interpolated process of the algorithm at the time instants $t_n$ where $t_n = \sum_{i=0}^{n-1} a(i)$. then define $X^n(t) = X^0(t+t_n)$. Then a book claims that $X^n(.)$ is equicontinuous. For that I need $X^0(.)$ to be uniformly continuous. For that I need $\lim_{t\rightarrow \infty} X^0(t)$ to exist. For that I need $x_n$ to converge. But $x_n$ does not converge. Then how to prove the equicontinuity
 A: I'm gonna make some assumptions, suggested in the comments:
i): $a_n \downarrow 0$ => $a_n$ positive,  amongst other things. 
I'll prove something first: if  $\sum a_n $ diverges to $+\infty$, then $(x_n)$ doesn't converge.
Indeed, if $(x_n) \rightarrow x $ ; $(h(x_n)) \rightarrow h(x)  $ (since h is continuous).
We get : $ x_{n+1} -x_n $ ~ $a_n*h(x) $
You can sum partial sums of positive terms  and you get :
$ x_n -x_1 $ ~ $h(x)*A_n$ ; where $A_n$ is the partial sum of the $a_k$. So $(x_n)$ diverges, and we get an absurdity. We can now assume that $\sum a_n $ converges (positive terms, either diverges to $+\infty$
 or converges).
$|x_{n+1}-x_n| = a_n*|h(x_n)|$ . $(x_n)$ bounded $\implies$  $(h(x_n))$ bounded 
=> $|x_{n+1}-x_n| \leq a_n*H$
But $\sum a_n$ converges, then so does $\sum |x_{n+1}-x_n|$. Let $v_n = x_{n+1}-x_n$. here we have proved that $\sum v_n$ converges absolutely, so $\sum v_n$ converges ie : 
$ V_n = \sum_{k=1}^n (x_{k+1}-x_k) = x_{n+1} - x_1 $ converges. Hence $(x_n)$ converges. 
