Convergence in a complete metric space

Question

I have two related questions. The first one is, show that if $(X, \rho)$ is a complete metric space and $\{ x_{n} \}$ is a sequence satisfying $\rho(x_{n}, x_{n+1}) < 2^{-n}$, then $\{ x_{n} \}$ converges. The second one is to find an example of a sequence $\{ x_{n} \}$ in a complete metric space such that $\rho(x_{n}, x_{n+1}) < 1/n$ and yet $\{ x_{n} \}$ does not converge.

I am thinking that the reason these two facts are true is something similar to the reason that the harmonic series $ \sum_{n=1}^{\infty}\frac{1}{n} $ does not converge but $ \sum_{n=1}^{\infty}\frac{1}{2^{n}} $ does converge. However, I cannot see how to prove the first fact or come up with a counterexample for the second fact.

Any help would be appreciated.

Answer 1

For the first problem, you need to show $(x_n)$ is Cauchy, so fix $\epsilon > 0$, then note that if $n>m$, $$ \rho(x_n,x_m) \leq \sum_{j=m}^n \frac{1}{2^j} \leq \sum_{j=m}^{\infty} \frac{1}{2^j} \qquad (\ast) $$ Being the tail of a convergent series, for $\epsilon > 0, \exists N\in \mathbb{N}$ such that $$ \sum_{j=N}^{\infty} \frac{1}{2^j} < \epsilon $$ From this and $(\ast)$, it follows that $(x_n)$ is Cauchy and hence convergent.

For the second problem, I will just give a hint: You mention the series $$ \sum_{n=1}^{\infty} \frac{1}{n} $$ Can you use this very example to find what you want in $\mathbb{R}$ (which is complete with respect the usual metric)?