I am attempting to solve the following:
Let $0<\alpha<1$. Suppose that $\{x_{n}\}$ is a sequence in a complete metric space $(X,\rho)$, and for each $n$, $\rho (x_{n},x_{n+1}) \leq \alpha^{n}$. Show that $\{ x_{n}\}$ converges. Does $\{x_{n}\}$ necessarily converge if we only require that for each $n$, $\rho (x_{n},x_{n+1}) \leq 1/n$?
Right now, I am working on the first part, showing that $\{x_{n}\}$ converges, and I am stuck.
Since $(X,\rho)$ is a complete metric space, I am trying to show that $\{x_{n}\}$ is Cauchy. I.e., I need to show that $\forall \epsilon > 0$, $\exists N \in \mathbb{N}$ for which if $n$, $m$ $\geq N$, then $\rho(x_{n},x_{m})<\epsilon$.
Also, since the triangle inequality seems to be a good friend of ours when working in metric spaces, I decided to try the following:
$\rho (x_{n}, x_{m}) \geq \rho(x_{n}, x_{n+1}) + \rho (x_{n+1},x_{m}) \leq \alpha^{n} + \rho (x_{n+1}, x_{m})$.
Now, I know that for $n$ sufficiently large, $\alpha^{n}$ goes to $0$, but I'm not sure how to use that to help me here. Also, I don't know how to put a bound on $\rho(x_{n+1}, x_{m})$ so that I can make the whole thing less than $\epsilon$. Please help.
Also, I'm not really sure at all what to do for the next part. I'd like to say the answer is yes, because sometimes I've seen limit problems where we use $1/n$'s instead of $\epsilon$'s, and I'd think that for $n$ sufficiently large, we would get that $\rho (x_{n},x_{n+1}) < \epsilon$, but also most of the time when a "Does the blah blah necessarily blah blah" is attached to problems, the answer is no! Any assistance you could offer would be much appreciated, as long as it isn't too cryptic/Socratic. Thank you.
Your first idea was good! You only need to go further. Explicitely, $$\rho(x_n,x_m)\leq \rho(x_n,x_{n+1})+\rho(x_{n+1},x_m)\leq ... \\ \leq \rho(x_n,x_{n+1})+ \rho(x_{n+1},x_{n+2})+...+\rho(x_{m-1},x_m)\leq \alpha^n+\alpha^{n+1}+...+\alpha^{m-1}.$$ Now you certainly now how to compute $\alpha^n+\alpha^{n+1}+...+\alpha^{m-1}$, and you just have to show that it is arbitrary small.
The same type of argument will give you the answer for the second part.