Triangle Inequality for proving Cauchy Criterion for Series I need to show that the statement $$\forall \epsilon > 0, \text{there is an } N\in \mathbb{N} \text{ such that for all } n\geq N, |\Sigma _{k=n+1}^{\infty} a_k | < \epsilon$$ 
implies $$\forall \epsilon > 0, \text{there is an } N\in \mathbb{N}, \text{such that if } n,m \geq N, |\Sigma_{k=n+1}^{m} a_k | < \epsilon $$
I understand intuitively why these statements would be equivalent. However, I'm having a hard time following the proof supplied in my textbook (Real Analysis and Applications by Davidson, Donsig). The proof begins like this: 

I don't understand how or why the triangle inequality implies:  $$|\Sigma_{k=n+1}^{m} a_k | \leq ||\Sigma_{k=n+1}^{\infty} a_k| - |\Sigma_{k=m+1}^{\infty} a_k||$$
 A: First, this is the reverse triangle inequality (not the triangle inequality) which states
$$ |x-y| \geq ||x| - |y||$$
Additionally, they are using 
\begin{align}
\sum_{k=n+1}^\infty a_k &= \sum_{k=n+1}^m a_k + \sum_{k=m+1}^\infty a_k\\
\sum_{k=n+1}^\infty a_k - \sum_{k=m+1}^\infty a_k&= \sum_{k=n+1}^m a_k\\
\end{align}
Applying the reverse triangle inequality we have
\begin{align}
|\sum_{k=n+1}^m a_k| &= |\sum_{k=n+1}^\infty a_k - \sum_{k=m+1}^\infty a_k|\\
&\geq ||\sum_{k=n+1}^\infty a_k| - |\sum_{k=m+1}^\infty a_k||
\end{align}
But this does not match with what they are saying, it seems they have the inequality backwards?
The following is adapted from a (different?) version of the book. If (2) holds then for $L \in \mathbb{R}$ we have for all $\varepsilon > 0$ there exists $N$ such that for $n \geq N$.
$$|L - \sum_{k=n+1}^\infty a_k| < \dfrac{\varepsilon}{2}$$
Then by the triangle inequality, for $m \geq N$ we have
\begin{align}
|\sum_{k=n+1}^m a_k| &= |\sum_{k=n+1}^\infty a_k - \sum_{k=m+1}^\infty a_k|\\
&= |\sum_{k=n+1}^\infty a_k - L + L - \sum_{k=m+1}^\infty a_k|\\
&\leq |\sum_{k=n+1}^\infty a_k - L| + |L - \sum_{k=m+1}^\infty a_k|\\
&=|L - \sum_{k=n+1}^\infty a_k| + |L - \sum_{k=m+1}^\infty a_k|\\
&< \dfrac{\varepsilon}{2} +  \dfrac{\varepsilon}{2}\\
&= \varepsilon
\end{align}
Thus $|\sum_{k=n+1}^m a_k| < \varepsilon$ as desired.
