Interpretting geometrically error bounds for alternate series 
Is there any way to interpret geometrically the bound errors of alternate series better?

I've been told that the error bound for an alternate series $\sum_{n=0}^\infty (-1)^n a_n$ is given by $$|R_n(x)|≤a_{n+1}$$ being $|R_n(x)|$ the error which also is given to be $S-S_n$? Not sure, but on any case I'm struggling to see how the error bound of alternate series is smaller than the next term of the sequence $a_n$, which is $a_{n+1}$ , so I'm wondering if anyone can provide a graph/drawing or whatever where I can see such thing clearer, because the way I was told this had no geometric inputs and I was looking for some and on internet I didn't quite found it either.
 A: This is intended to give a graphic intuition of the result and is by no means a rigourous proof ; mainly, we won't try to understand why the series we manipulate are convergent.
Let's note $u_n=(-1)^n a_n$. We can represent this sequence with the following graph :

For $n \in \mathbb{N}$, the error is
$$R_{n} = \sum\limits_{k=0}^{+\infty} u_k - \sum\limits_{k=n}^{+\infty} u_k = \sum\limits_{k=n+1}^{+\infty} u_k.$$
On our graph, $R_n$ is the sum of all the bars starting at $u_{n+1}$ (the blue ones are positive, the red ones are negative) :

Let's first show that $R_n$ as the same sign as $u_{n+1}$ (which is $(-1)a_n$). As the sequence $(a_n)$ is decreasing (definition of an alternating series), the size of the bars in our graph is smaller and smaller. Thus, by adding two consecutive terms, we get a number that has the same sign as the first of the two terms. For example, if we have two consecutive terms were the first is positive, we get the following picture :

Thus, in this picture where $u_{n+1} \geqslant 0$, $R_n$ can be obtained by adding all the pairs of terms starting from $u_{n+1}$, i.e. by adding positive values : thus $R_n$ is positive. If we begun at a blue bar ( where $u_{n+1} \leqslant 0$, $R_n$ would be negative with the same argument. Thus $R_n$ has the same sign as $u_{n+1}$.
$\rhd$ Now, this implies that $R_{n+1}$ has the opposite sign of $R_n$, i.e. the opposite sign of $u_{n+1}$. Thus, as $R_{n} = u_{n+1}+ R_{n+1}$, the term $R_{n+1}$ will partially cancel with $u_{n+1}$ and will give $|R_n| \leq |u_{n+1}|$.

To be more precise, if $u_{n+1} \geq 0$, then we have seen previously that $R_n \geq 0$ and $R_{n+1} \leq 0$. Thus
$$ 0 \leq R_n  = u_{n+1} + R_{n+1} \leq u_{n+1}$$
and we get $|R_n| \leq |u_{n+1}|$. A similar proof holds true in the case $u_{n+1} \leq 0$. This is the inequality $|R_n| \leq a_{n+1}$ you were looking for.
A: Draw a number line.
Consider the series $a_0-a_1+a_2-a_3+a_4+ ...$ with $a_k$ positive and decreasing.
The partial sums are $S_0 = a_0$, $S_1 = a_0-a_1$, $S_2 = a_0-a_1+a_2$, etc.
The way you obtain the partial sums on the number line is by moving $a_0$ to the right, then $a_1$ to the left, then $a_2$ to the right, then $a_3$ to the left, etc.
Notice that  the sequence of intervals $I_n = [\textrm{min}(S_n), \textrm{max}(S_{n+1})]$ satisfies $I_{n+1} \subset I_{n}$ and the length of $I_n$ is $|a_{n+1}|$.  The intersection of all of these intervals is the infinite sum $S$.  We can see that since $S \in I_{n}$ we have $|S - a_n|< |a_{n+1}|$.
