Absolute convergence in a metric space Let $(X,d)$ be a metric space, $(a_n)$ and $(b_n)$ are sequences in $(X,d)$. If $\sum_{n=1}^\infty d(a_n,b_n)$  is absolutely convergent, what do I say about the convergence of $(a_n)$ and  $(b_n)$?
 A: Take $(\Bbb R,|.|)$ the ususal metric on $\Bbb R$ and $a_n=n,b_n=n+\frac1{n^2}$ so what we can conclude?
A: If $\sum_{n=1}^{\infty}d(a_n, b_n)$ is absolutely convergent (which is equivalent to convergent because $d(a_n, b_n) \geq 0$), then $\lim\limits_{n\to\infty} d(a_n, b_n) = 0$.
Suppose $a_n$ converges to $a$, then I claim that $b_n$ converges to $a$ as well. Note that we have
$$d(a_n, b_n) \geq |d(a_n, a) - d(b_n, a)|$$ by the reverse triangle inequality. As $n \to \infty$, $\lim\limits_{n\to\infty}|d(a_n, a) - d(b_n, a)| = 0$. 
As $a$ was arbitrary, we can state that under the given conditions, $\{a_n\}$ converges if and only if $\{b_n\}$ converges, in which case, they both converge to the same limit. If they don't both converge, they both diverge, but in such a way that for any $a \in X$, $\lim\limits_{n\to\infty}|d(a_n, a) - d(b_n, a)| = 0$.
A: Nothing, because $a_n$ could be any sequence and $b_n$ could equal $a_n$.
A: If the series converges, then $d(a_n,b_n)\to 0$, and so if $a_n\to L$, then since $d(b_n,L)\le d(a_n,b_n)+d(a_n,L)$, it follows that $b_n\to L$. In other words, the sequences converge and diverge together, and necessarily to the same limit. However, this only uses the convergence $d(a_n,b_n)\to 0$ and not the convergence of the series, which is much stronger. The convergence of the series implies that when $a_n$ converges, the speed at which $b_n$ converges to the common limit can't be asymptotically a lot slower than the convergence of $a_n$ (however this is hardly a precise statement, more of a heuristic that can be turned more precise). I don't see what else the convergence of the series yields.  
