---
tag: KAX5R5F099
kind: proof
name: Proof of the divergence part of the ratio test
msc: [40A05]
proves: 08AM0KRCZ1
uses: [0XKWSSY8P2, 0VVY4H92G2]
sources:
  - url: https://en.wikipedia.org/wiki/Ratio_test
    locator: Proof
---

## Proof

1. Apply the definition of the [[limit|0XKWSSY8P2]] with $\varepsilon = L - 1 > 0$. There is $N$ with $\abs{a_{n+1}}/\abs{a_n} > L - \varepsilon = 1$ for all $n \ge N$.
2. So $\abs{a_{n+1}} > \abs{a_n}$ for $n \ge N$. By induction, $\abs{a_n} \ge \abs{a_N}$ for all $n \ge N$.
3. By hypothesis $a_N \ne 0$, so $\delta = \abs{a_N} > 0$.
4. If $a_n \to 0$, then $\abs{a_n} < \delta$ for all large $n$. This contradicts step 2.
5. So $(a_n)$ does not converge to $0$. By the [[term test|0VVY4H92G2]], the series $\sum a_n$ diverges.

## Idea

When the ratio tends to a number above $1$, the absolute values of the terms eventually grow. Growing nonzero terms cannot tend to $0$, so the series fails the term test.