Q:

Explanation regarding alternating sum estimations

Can somebody please explain why $|s_{n+1}-s_n |\geq |s-s_n|$?

Error estimate for alternating series
If the sequence $\{a_n\}$ satisfies the conditions of the Alternating Series Test so that the series $\sum_{n=1}^\infty a_n$ converges to the sum $s$, then the error in the approximation $s\approx s_n$ (where $n\ge N$) has the same sign as the first omitted term $a_{n+1} = s_{n+1} - s_n$, and its size is no greater than the size of that term: $$|s-s_n| \le |s_{n+1}-s_n| = |a_{n+1}|$$

My confusion
$s=\sum_1^\infty a_n $, $s_{n+1} = \sum^{n+1}_1 a_n$. So isnt by definition $s>s_{n+1}$ and hence $|s_{n+1}-s_n |\leq |s-s_n|$?

A:

Using that $|a_n|$ is decreasing, can you show that $s\in [s_n;s_{n+1}]$ when $a_{n+1}>0$ and $s\in [s_{n+1};s_n]$ when $a_{n+1}<0$ ?
edit:
That means, can you show that $a_{n+1}>0\Rightarrow s_n<s<s_{n+1}$ and that $a_{n+1}<0\Rightarrow s_{n+1}<s<s_n$ ?
second edit:
The trick in that kind of proof is to consider the two monotonous series $s_{2n}$ and $s_{2n+1}$. One is increasing (wlog, let's say the first one) and the other one is decreasing, but both have the same limit $s$. So you can conclude that $\forall n,s>s_{2n}$ and $s<s_{2n+1}$.
third edit:
Assume $a_1>0$. 
Because $a_n$ is an alternating series, $a_{2n+1}>0$ and $a_{2n}<0,\forall n$. 
Because $a_n$ match the convergence criteria, $|a_n|$ is decreasing to zero. Notice that the subseries $|a_{2n}|$ and $|a_{2n+1}|$ are also decreasing and convergent to $0$.
For any $n$, $s_{2(n+1)}-s_{2n}=a_{2n+1}+a_{2n+2}=|a_{2n+1}|-|a_{2n+2}|>0$ because $a_{2n+1}>0$ and $a_{2n+2}<0$, and because $|a_n|$ is decreasing.
We conclude that $s_{2n}$ is increasing. As a subseries of $s_n$, it converges to $s$, so $\forall n, s_{2n}<s$.
The same reasonning with the series $s_{2n+1}$ proves that $\forall n, s<s_{2n+1}$.
From $\forall n, s_{2n}<s<s_{2n+1}$, we conclude that $|s_{2n+1}-s|<|s_{2n+1}-s_{2n}|$.
From $\forall n, s_{2n-1}>s>s_{2n}$, we conclude that $|s_{2n}-s|<|s_{2n}-s_{2n-1}|$.
Since this covers both possible parity of the indice, we have $\forall n,|s_{n+1}−s_n|≥|s−s_n|$

