First we show that for each $n > m$, we have $0 \le a_{m + 1} - a_{m + 2} + a_{m + 3} - \dotsb \pm a_n \le a_{m + 1}$. This will be achieved by means of the Second Principle of Mathematical Induction. For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :$0 \le a_{m + 1} - a_{m + 2} + a_{m + 3} - \dotsb \pm a_n \le a_{m + 1}$ === Basis for the Induction === $\map P 1$ holds, as $a_{m + 1} \ge 0$ by definition and $a_{m + 1} \le a_{m + 1}$. $\map P 2$ also holds, as $a_{m + 2} \le a_{m + 1}$ and so: :$0 \le a_{m + 1} - a_{m + 2} \le a_{m + 1}$ This is the basis for the induction. === Induction Hypothesis === In order to simplify the algebra, let: :$b_k := a_{m + 1} - a_{m + 2} + a_{m + 3} - \dotsb \pm a_k$ The $\pm$ signifies the fact that $a_k$ is positive for $k$ odd and negative for $k$ even. Suppose that $\forall j \le k: \map P j$ holds: :$0 \le b_j \le a_{m + 1}$ This is the induction hypothesis. We now show that $\forall k: \map P k \implies \map P {k + 1}$. === Induction Step === This is the induction step: Suppose $k$ is odd. By the induction hypothesis: :$0 \le b_k \le a_{m + 1}$ Because $\map P k$ holds: :$0 \le b_{k - 1} + a_k \le a_{m + 1}$ But as $a_k \ge a_{k + 1}$: :$a_k - a_{k + 1} \ge 0$ and so: :$0 \le b_{k - 1} + \paren {a_k - a_{k + 1} } = b_{k + 1}$ But as $b_k \le a_{m + 1}$: :$b_k - a_{k + 1} = b_{k + 1} \le a_{m + 1}$ So: :$0 \le b_{k + 1} \le a_{m + 1}$ or: :$0 \le a_{m + 1} - a_{m + 2} + a_{m + 3} - \dotsb - a_{k + 1} \le a_{m + 1}$ Thus for odd $k$ it follows that $\map P k \implies \map P {k + 1}$. Now suppose $k$ is even. By the induction hypothesis: :$0 \le b_k \le a_{m + 1}$ Then: :$0 \le b_k + a_{k + 1} = b_{k + 1}$ Because $\map P k$ holds: :$0 \le b_{k - 1} - a_k \le a_{m + 1}$ But as $a_k \ge a_{k + 1}$: :$a_k - a_{k + 1} \ge 0$ and so: :$b_{k + 1} = b_{k - 1} - a_k + a_{k + 1} =  b_{k - 1} - \paren {a_k - a_{k + 1} } = b_{k + 1} \le a_{m + 1}$ So: :$0 \le b_{k + 1} \le a_{m + 1}$ or: :$0 \le a_{m + 1} - a_{m + 2} + a_{m + 3} - \cdots + a_{k + 1} \le a_{m + 1}$ Thus for even $k$ it follows that $\map P k \implies \map P {k + 1}$. So for both even and odd $k$ it follows that $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction. Therefore for each $n > m$, we have: :$0 \le a_{m + 1} - a_{m + 2} + a_{m + 3} - \dotsb \pm a_n \le a_{m + 1}$ Now, let $\sequence {s_n}$ be the [[Definition:Series|sequence of partial sums of the series: :$\displaystyle \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} a_n$]] Let $\epsilon > 0$. Since $a_n \to 0$ as $n \to \infty$: :$\exists N: \forall n > N: a_n < \epsilon$ But $\forall n > m > N$, we have: {{begin-eqn}} {{eqn | l = \sequence {s_n - s_m}       | r = \size {\paren {a_1 - a_2 + a_3 - \dotsb \pm a_n} - \paren {a_1 - a_2 + a_3 - \dotsb \pm a_m} }       | c =  }} {{eqn | r = \size {\paren {a_{m + 1} - a_{m + 2} + a_{m + 3} - \dotsc \pm a_n} }       | c =  }} {{eqn | o = \le       | r = a_{m + 1}       | c = from the above }} {{eqn | o = <       | r = \epsilon       | c = as $m + 1 > N$ }} {{end-eqn}} Thus we have shown that $\sequence {s_n}$ is a Cauchy sequence. The result follows from Convergent Sequence is Cauchy Sequence. {{qed}}