**Lemma** Suppose that $(a_{n})$ is a [[Convergence|sequence]] of positive terms and that $\frac{a_{n+1}}{a_{n}} \to r$as $n \to \infty$ then: 1. If $r<1$ then $a_{n} \to 0$ 2. If $r>1$ then $a_{n} \to \infty$. Note that this generalises [[Criterion for Convergence of Geometric Sequence]]. ###### **Proof** Suppose that $r<1$. Choose $\epsilon>0$ such that $r+\epsilon<1$, there exists $N$ such that $\frac{a_{n+1}}{a_{n}}<r+\epsilon$for all $n \geq N$. So $0\leq a_{N+k} \leq (r+\epsilon)^{k}a_{N}$ for all $k \geq 1$. Since $|r+\epsilon|<1$, $(r+\epsilon)^{k} \to 0$; so $a_{N+k} \to 0$ as $k \to \infty$ using [[Sandwich Rule]], which shows that $a_{n} \to 0$ as $n \to \infty$ (by [[Shift Rule for Limits]]). On the other hand if $r>1$, choose $\epsilon>0$ such that $r-\epsilon>1$; there exists $N$ such that $\frac{a_{n+1}}{a_{n}} > r- \epsilon \quad \text{for all }n\geq N$and $a_{N+k}>(r-\epsilon)^{k}a_{N}$ for all $k \geq 1$. Since $r- \epsilon>1$, $(r-\epsilon)^{k} \to \infty$ as $k \to \infty$ so $a_{N+k} \to \infty$ as $k \to \infty$ so by [[Comparison Test for Sequences]] $a_{n} \to \infty$ as $n \to \infty$. # Applications