Prove Alternating Series Approximation Prove if $S=\sum_{n=1}^{\infty}a_{n}$ is an alternating series with $\left | a_{n+1}\right | < \left | a_{n} \right |$, and $\lim_{n\to\infty}a_{n}=0$, then $\left |S-(a_{1}+a_{2}+\cdots+a_{n}) \right | \leq \left | a_{n+1} \right | $ I'm supposed to group the terms in the error as $(a_{n+1}+a_{n+2})+(a_{n+3}+a_{n+4})$ to show that the error has the same sign as $a_{n+1}$ but I don't understand what they mean by the terms in the error. 
 A: It is I think best to look at a particular example, or two, or three. Consider the alternating series
$$1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23}+\cdots.\tag{1}$$
This is the famous Leibniz series. It converges to $\frac{\pi}{4}$. Here $a_n=\frac{(-1)^{n-1}}{4n-3}$.
Suppose that we truncate this series just after the term $-\frac{1}{7}$. Then the truncation error (in the language of your question, the sum of the terms in the error) is the rest of the stuff, which is
$$\frac{1}{9}-\frac{1}{11}+\frac{1}{13}-\frac{1}{15}+\frac{1}{17}-\frac{1}{19}+\frac{1}{21}-\frac{1}{23}+\cdots.\tag{2}$$
The expression (2) may be rewritten as 
$$\left(\frac{1}{9}-\frac{1}{11}\right)+\left(\frac{1}{13}-\frac{1}{15}\right)+\left(\frac{1}{17}-\frac{1}{19}\right)+\left(\frac{1}{21}-\frac{1}{23}\right)+\cdots.\tag{3}$$
It is also useful to group a little differently, as
$$\frac{1}{9}+\left(-\frac{1}{11}+\frac{1}{13}\right)+\left(-\frac{1}{15}+\frac{1}{17}\right)+\left(-\frac{1}{19}+\frac{1}{21}\right)+\cdots.\tag{3'}$$
Note that the sum of any two grouped terms in (3) is positive. So the sum $S$ of the full series (2) is positive. The estimate $1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}$ is an underestimate of $S$. 
Now look at the version (3') of the truncation error (2). The sum of any two grouped terms is negative, so the full sum is less than $\frac{1}{9}$.
We conclude that the error when we truncate just after the $-\frac{1}{7}$ term is positive but less than $\frac{1}{9}$. The absolute value of the error is less than the absolute value of the first neglected term.
I suggest you make a similar analysis of the truncation error if we truncate just after the $\frac{1}{9}$ term. Grouping in the two ways done above will show that the sum of the neglected terms is negative, but greater than $-\frac{1}{11}$. Again the truncation error is less in absolute value than the absolute value of the first neglected term.
We can make an essentially identical analysis for general series $a_1+a_2+a_3+a_4+\cdots$, where the terms alternate in sign, decrease in absolute value, and have limit $0%
The truncation error is always less in absolute value than the absolute value of the  first neglected term. If the first neglected term is positive, then our estimate of the full infinite sum  is an underestimate. If the first neglected term is negative, our estimate is an overestimate. 
A: assume we have a convergent alternating series $$s_\infty = a_1 - a_2 + a_3 +\cdots $$ we write $$s_\infty -(a_1 - a_2 +\cdots + (-1)^{n-1}a_n = error =  (-1)^n\left(a_{n+1} - a_{n+2}+\cdots \right)$$  
let us look at the case  $n$ even.  then using the fact that $a_{j+1} \le a_j, \forall j$
$$\begin{align}error & = a_{n+1} - a_{n+2}+a_{n+1} + \cdots\\
&=(a_{n+1} -a_{n+2}) + (a_{n+3}-a_{n+4}) + \cdots \ge 0\\
&=a_{n+1}-(a_{n+2} - a_{n+3})+(a_{n+4}-a_{n+5}+\cdots ) \le a_n\end{align}$$ 
the grouping of the error in two ways shows that $$0 \le error \le a_{n+1}, n \, \text{ even. }$$ in a similar way you can show that $$-a_{n+1} \le error \le 0, n \, \text{ odd. }$$
the two can be combined to state the fact that the error has the same sign as the first omitted tern and is smaller in magnitude than the same.
