**Lemma** Suppose that$|a_{n}|^ {\frac{1}{n}} \to r \quad \text{as} \quad n\to \infty$then: 1. if $r<1$ then $\sum a_{n}$ [[Absolutely Convergent Series|converges absolutely]]. 2. if $r>1$ then $\sum a_{n} = \infty$ **Proof** **(similar to that of [[Ratio Test for Series]])** 1. Suppose $r<1$. Take $\epsilon>0$ such that $\rho:=r+\epsilon<1$ there exists $N$ such that for all $n \geq N$ we have $|a_{n}|^{\frac{1}{n}} < \rho \implies |a_{n}|< \rho^{n}$Since $\rho<1$ the geometric series $\sum \rho^{n}< \infty$ so $\sum |a_{n}|$ converges by [[Comparison Test for Series With Non-Negative Terms]]. 2. If $r>1$ we take $\epsilon>0$ such that $p:=r-\epsilon>1$; then there exists $N$ such that for all $n\geq N$ we have $|a_{n}|^{\frac{1}{n}} > \rho \implies |a_{n}| > \rho^{n}>1 $in particular $a_{n} \not \to 0$, so $\sum a_{n}$ cannot converge.