Why does the series $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}$ converge? I saw the proof, that the alternating harmonic series converges. So $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$ converges. I didn't understand one thing in proof. So the argument here is, if we look at subsequence
$s_{2n}=(\frac{1}{1}-\frac{1}{2})+(\frac{1}{3}-\frac{1}{4})+\ldots+ (\frac{1}{2n-1}-\frac{1}{2n})$ we can see, that this subsequence is monotone increasing and if we look at subsequence
$s_{2n+1}=\frac{1}{1}-(\frac{1}{2}-\frac{1}{3})-(\frac{1}{4}-\frac{1}{5})-\ldots-(\frac{1}{2n}-\frac{1}{2n+1})$ is monotone decreasing. And since
$0<s_{2n}<s_{2n}+\frac{1}{2n+1}=s_{2n+1}\leq1$ we have that both of these subsequences are bounded
below by 0 and above by 1. Therefore they are both convergent and to the same value. Why does it follow? From which theorem? So I can't understand it. Thank you in advance!
 A: There are two theorems at work here:
Theorem 1: A bounded, monotonic sequence must converge
This is why both $(s_{2n})$ and $(s_{2n+1})$ are convergent sequences.
Theorem 2: If $a_n \leq b_n \leq c_n$ for all $n$ and $\lim_{n \to \infty} a_n = \lim_{n \to \infty} c_n = L$, then $b_n$ converges with $\lim_{n \to \infty} = L$.
To use this second result, we note that
$$
\lim_{n \to \infty} s_{2n} + \frac 1{2n+1} = \lim_{n \to \infty} s_{2n} + \lim_{n \to \infty} \frac 1{2n+1} = \lim_{n \to \infty} s_{2n} + 0 = \lim_{n \to \infty} s_{2n},
$$
so taking $a_n = s_{2n}$ and $c_n = s_{2n} + \frac 1{2n+1}$, we can conclude that $\lim_{n \to \infty} s_{2n} = \lim_{n \to \infty} s_{2n+1}$
A: This does not answer your question, but I'll write the answer anyway so you can look at the problem from a different point of view.
Using Taylor series of $\log (1+x)$, we have:
$$\log (1+x)=\sum_{n=1}^{+\infty} \frac{(-1)^{n+1}x^n}{n} \quad \text{for} \quad |x|\leq1, x\neq-1 $$
$$\sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}{n}=\sum_{n=1}^{+\infty} \frac{(-1)^{n+1}1^n}{n}=\log 2 $$
