**Lemma** Suppose that $a_{n} \geq 0$ with $a_{n+1} \leq a_{n}$ and $a_{n} \to 0$. Then $\sum_{n=1}^{\infty} (-1)^{n+1} a_{n} < \infty$ **Generalises** [[Alternating Harmonic Series]]. **Proof** Let $s_{k}$ denote the partial sums. We have that $s_{2k} \leq s_{2k+1}$, since $s_{2k+1} = s_{2k} + a_{2k+1}$. Notice that $s_{2k+2} - s_{2k} = a_{2k+1}- a_{2k+2} \geq 0$so $(s_{2k})$ is an increasing sequence. Similarly $s_{2k+1} - s_{2k-1} = -a_{2k} + a_{2k+2} \leq 0$so $(s_{2k+1})$ is a decreasing sequence. In particular $s_{2k+1} \leq s_{1}$ for every $k$. It follows that $s_{2k} \leq s_{2k+1} \leq s_{1}$. Since $(s_{2k})$ is increasing and bounded above, $s_{2k} \to l \in \mathbb{R}$ by [[Monotone Bounded Real Sequence is Convergent|monotone convergence]]. Since $s_{2k+1} =s_{2k}+a_{2k+1}$ and $a_{2k+1} \to 0$ as $k \to \infty$, by [[Algebra of Limits of Convergent Sequences|sum rule]], $s_{2k+1} \to l$ also. By [[If even and odd terms of sequence converge to the same limit then the sequence converges to that limit]], $s_{k} \to l$ as $s_{k} \to \infty$. **Summary** - We use the fact that $(a_{n})$ is **non-negative** to show that note that $s_{2k} \leq s_{2k+1}$. - We use the fact that $(a_{n})$ is **decreasing** to show that and $s_{2k}$ is increasing while $s_{2k+1}$ is decreasing and bounded by $s_{1}$ hence $s_{2k}$ is also bounded by $s_{1}$. - Since $s_{2k}$ is increasing and bounded above, it is convergent. - We use the fact that $a_{n}$ **converges to 0** and sum rule to show that $s_{2k+1}$ converges to the same limit. - So $s_{k}$ converges to the same limit.