**Lemma** If $a_{n} \neq 0$ and $\left| \frac{a_{n+1}}{a_{n}} \right| \to r \quad as \quad n \to \infty$then 1. if $0\leq r <1$ then $\sum a_{n}$ [[Absolutely Convergent Series|converges absolutely]] 2. if $r>1$ then $\sum a_{n}$ does not converge 3. if $r=1$ it could be either **Proof** 1. Suppose $0\leq r<1$. Choose $\epsilon>0$ such that $\rho:= r+\epsilon <1$. So $\exists N \in \mathbb{N}$ such that for all $n \geq N$, $\left| \frac{a_{n+1}}{a_{n}} -r\right| < \epsilon $So $-r-\epsilon < r-\epsilon < \frac{a_{n+1}}{a_{n}} < r+ \epsilon \implies \left| \frac{a_{n+1}}{a_{n}} \right| < \rho .$Then for all $n \geq N+1$, $|a_{n}| \leq \rho^{n-N} |a_{N}|$and it follows that $\sum a_{n}$ converges absolutely by [[Comparison Test for Series using Absolute Value of Terms (Corollary 2)|comparison test]] since for $\rho<1$ the geometric series $\sum \rho^{n-N} |a_{N}|$ converges. 2. If $r > 1$, we take $\epsilon>0$ such that $1<\rho:= r-\epsilon<r$ there exists $N$ such that $ \left| \frac{a_{n+1}}{ a_{n}} \right| > \rho \quad \forall n\geq N$Then $|a_{n}| > \rho^{n-N} |a_{N}| \geq |a_{N}|$ for all $\geq N$. So $a_{n} \not \to 0$ hence $\sum a_{n}$ does not converge since [[Terms of Convergent Series Tend to Zero]]. 3. $\sum \frac{1}{n}$ and $\sum \frac{1}{n^2}$ show that we cannot decide if $r=1$.