Let $\set{p_{n}}$ be a positive [[Series|Infinite Series]] that [[Convergent Series|Converges]] to $0$. The order of convergence $\alpha \in [1,\infty)$ for the series $p$ is defined as: $\huge \alpha = \lim_{ n \to \infty } \frac{\ln p_{n+1}}{\ln p_{n}} $ ### Arithmetic Relations When analyzing the order of convergence for the sum of two [[Series]] $p_{n},q_{n}$ the order is determined by the **minimum** of their individual orders. $\huge \alpha_{p+q} = \min(\alpha_{p},\alpha_{q}) $ THis is because the lower order will 'slow down' the convergence. However, if we take the product of the two series, then their order is determined by the maximum of their individual orders. $\huge \alpha_{p\cdot q} = \max(\alpha_{p},\alpha_{q}) $ ### Recurrence Relation Property $\huge \forall k <\alpha: \lim_{ n \to \infty } \frac{p_{n+1}}{p_{n}^{\alpha}} = 0 $ $\huge \forall k >\alpha: \lim_{ n \to \infty } \frac{p_{n+1}}{p_{n}^{\alpha}} = \infty $ Note that if $k=\alpha$ that means that the given limit could either be [[Convergent Series|Convergent]] OR [[Divergence|Divergent]]. >[!example]- >Let $\set{p_{n}}$ be a sequence that converges to $0$ such that: >$ \huge p_{n+1} = 3p_{n}^{2} $ > >$\huge \begin{align} >\lim_{ n \to \infty } \frac{p_{n+1}}{p_{n}^{2}} = 3 >\end{align}$ >Therefore $\alpha=2$.