Does $d(x_{n+1},x_n)Let $(X,d)$ be a complete metric space, $(x_n)_{n\in\mathbb{N}}\subset X$ such that $d(x_{n+1},x_n)<d(x_n,x_{n-1})$ for all $n\in\mathbb{N}$. Since I cannot construct such sequence which is not convergent, I wonder if every such sequence is convergent? I first thought of defining a sequence $y_n:=d(x_{n+1},x_n)\in[0,\infty)$ and checking if it converges to zero, but it does not seem fruitful. (Of course, I am aware of the definition of convergence, which is much stronger than this.)
 A: The sequence of partial sums $S_n:=\sum_{i=1}^n1/i$ satisfies $d(S_{n+1},S_n)<d(S_n,S_{n-1})$ but is not convergent since $\lim_{n\to\infty}S_n=\sum_{i=1}^\infty1/i$ and the harmonic series does not converge.
A: The assumption given here means that the distance between consecutive terms of $x_n$ is strictly decreasing. Thus the distance between consecutive terms converges to a limit.
This does not mean that the sequence $x_n$ will converge, for example $d(x_n, x_{n-1})$ might tend to a nonzero limit as in the example given by vadim123. Even if $d(x_n, x_{n-1}) \rightarrow 0$, it still does not follow that $x_n$ has a limit, for a counterexample see the answer by gofvonx.
If you want convergence, you need stronger assumptions for the distance between consecutive terms. For example, the following is a common trick: if you assume that $d(x_n, x_{n-1}) \leq a_n$, where $\sum_{i = 0}^\infty a_i$ converges, it follows that $x_n$ is Cauchy, and thus converges if you assume completeness. 
A: Let $x_0=1$, $x_n=x_{n-1}+1+\frac{1}{n}$ for $n\ge 1$.  This satisfies your condition but is not convergent.
