A basic doubt on a infinite series problem I see in Rudin the following statement is claimed for the following convergent series: 
$$1-\frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \dots$$
If $s$ is the sum of this series then $$s \lt 1 -\frac{1}{2} + \frac{1}{3}$$. How is that possible to tell without knowing $s$ ?
 A: $$S_{2k+1} = 1-\frac{1}{2} + \frac{1}{3} - \left(\frac{1}{4} - \frac{1}{5}\right) - \ldots -\left(\frac{1}{2k} - \frac{1}{2k+1}\right)  < 1-\frac{1}{2} + \frac{1}{3}-\left(\frac{1}{4} - \frac{1}{5}\right)$$ and $S_{2k+1}\to s$ imply
$$s \leqslant 1-\frac{1}{2} + \frac{1}{3}-\left(\frac{1}{4} - \frac{1}{5}\right) < 1-\frac{1}{2} + \frac{1}{3}.$$
A: It's a known fact: suppose you have a decreasing sequence $(a_n)_{n\in\mathbb{N}}$  of positive numbers and that $\lim_{n\to\infty}a_n=0$. Then the series
$$
\sum_{n=0}^\infty (-1)^n a_n
$$
converges to some number $s$ and the following holds for any natural number $k$:
$$
\sum_{n=0}^{2k+1}a_n < s < \sum_{n=0}^{2k} a_n
$$
Let's prove, by induction on $k$, that the sequence $(s_{2k+1})_{k\in\mathbb{N}}$ is increasing and that $(s_{2k})_{k\in\mathbb{N}}$ is decreasing, where
$$
s_k=\sum_{n=0}^k (-1)^n a_n.
$$
Indeed
$$
s_{2k+3}-s_{2k+1}=-a_{2k+3}+a_{2k}>0
$$
and
$$
s_{2k+2}-s_{2k}=a_{2k+2}-a_{2k+1}<0.
$$
Therefore, if the alternating series converges to $s$ we must have the inequality above, because
$$
\lim_{k\to\infty}s_{2k+1}=s=\lim_{k\to\infty}s_{2k}.
$$
The fact that the alternating sequence converges stems from the fact that
$$
s_{2k+2}-s_{2k+1}=a_{2k+2}
$$
and so we can make this difference as small as we want, since the sequence $(a_n)$ converges to zero by hypothesis.
A: Notice that
$$ 1-\frac{1}{2}+\frac{1}{3}-\left( \frac{1}{4}-\dots \right)=1-\frac{1}{2}+\frac{1}{3}-a< 1-\frac{1}{2}+\frac{1}{3}$$
Since $a>0$.
Note:
$$a=\frac{1}{4}-\frac{1}{5}+\dots.$$
