# Proof

• 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 $$s_{2k}\ge s_1=a_0-a_1$$ 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 $L-U=\lim_{k\to\infty}(s_{2k}-s_{2k+1})=\lim_{k\to\infty} a_{2k+1}=0.$
• 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

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983