I want to show that $p_n$ a sequence in X converges if $\Sigma_{n=1}^{\infty} d(p_{n}, p_{n+1})$ converges. Here's the outline of what I've done so far:
If X is a complete metric space, then every Cauchy sequence converges. WTS that $p_n$ is a Cauchy sequence. By triangle inequality and letting m=n+k, we have that:
$d(p_n, p_m) = d(p_n, p_{n+k}) \leq d(p_n, p_{n+1}) + d(p_{n+1}, p_{n+2}) .... d(p_{n+k-1}, p_{n+k})$.
And I'm stuck. I know I by definition of Cauchy that I need $\epsilon$ such that $\forall \epsilon >0, \exists N \text{ such that } d(p_n, p_m)< \epsilon$.
Help would be much appreciated!
If $\Sigma_{n=1}^{\infty} d(p_{n}, p_{n+1})$ converges then the sequence of partial sums, $\{S_n\}$ is Cauchy where $$S_k=\Sigma_{n=1}^{k} d(p_{n}, p_{n+1})$$. Then for any given $\epsilon>0$,$\exists m_0\leq n_0\in\mathbb{N}$ such that $S_n-S_{m-1}<\epsilon$ for all $m\geq m_0, n\geq n_0\in\mathbb{N}$ with $m\leq n$. The rest you have done already!