Proof
(related to Proposition: Leibniz Criterion for Alternating Series)
 Let $(a_n)_{n\in\mathbb N}$ being a monotonically decreasing real series, which is convergent to \(0\), i.e. with \(\lim_{n\to\infty} a_n=0\).
 Define the partial sums $s_k:=\sum_{n=0}^k (1)^n a_n.$
 Then, for the partial sums \(s_k\) with even indices we have that $s_{2k+2}s_{2k}=a_{2k+1}+a_{2k+2}\le 0.$
 Therefore, $s_0\ge s_2\ge s_4\ge \ldots \ge s_{2k}\ge s_{2k+2}\ge\ldots.$
 Since \(s_{2k}\ge s_1=a_0a_1\) for all \(k\), the subsequence \((s_{2k})_{k\in\mathbb N}\) is monotonically decreasing and bounded.
 Analogously, for the partial sums \(s_k\) with odd indices $s_{2k+3}s_{2k+1}=a_{2k+2}a_{2k+3}\ge 0.$
 Therefore, $s_1\le s_3\le s_5\le \ldots \le s_{2k+1}\le s_{2k+3}\le\ldots.$
 Since \(s_{2k+1}\le s_0=a_0\) for all \(k\), the subsequence \((s_{2k+1})_{k\in\mathbb N}\) is monotonically increasing and bounded.
 Therefore, according to the monotone convergence theorem, both sequences converge to some limits $$\lim_{k\to\infty} s_{2k}=L\quad\text{and}\quad\lim_{k\to\infty} s_{2k+1}=U.\quad\quad( * )$$
 In fact, \(L=U\), since $LU=\lim_{k\to\infty}(s_{2k}s_{2k+1})=\lim_{k\to\infty} a_{2k+1}=0.$
 It remains to be shown that the common limit of the subsequences \((s_{2k})_{k\in\mathbb N}\) and \((s_{2k+1})_{k\in\mathbb N}\) is also a limit of the sequence \((s_{k})_{k\in\mathbb N}\) itself.
 Choose \(\epsilon > 0\) arbitrarily small, but fixed.
 Then, from \( ( * ) \), it follows that there exist some indices \(N_1(\epsilon),N_2(\epsilon)\in\mathbb N\) with $s_{2k}  L < \epsilon$ for all $k\ge N_1(\epsilon),$ and $s_{2k+1}U < \epsilon$ for all $k\ge N_2(\epsilon).$
 It follows $s_{k}  L < \epsilon$ for all $k\ge \max(N_1(\epsilon),N_2(\epsilon)).$
 This proves that the alternating series $\sum_{n=0}^\infty (1)^n a_n$ is convergent.
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References
Bibliography
 Forster Otto: "Analysis 1, Differential und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983