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: $$|\Sigma_{k=n+1}^{m} a_k | \leq ||\Sigma_{k=n+1}^{\infty} a_k| - |\Sigma_{k=m+1}^{\infty} a_k||$$
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.