---
tag: T13BF143VR
kind: proof
name: Proof of the ratio test
msc: [40A05]
proves: 0QNY6ZBXPN
uses: [0W3K3QPEWT, 0XKWSSY8P2, 0N32V0E3XJ, 05APWSX53S, 0MX1HZBBXJ]
sources:
  - url: https://en.wikipedia.org/wiki/Ratio_test
    locator: Proof
---

## Proof

We compare $\abs{a_n}$ with a geometric sequence.

1. Each ratio $\abs{a_{n+1}}/\abs{a_n}$ is at least $0$. Since [[limits preserve non-strict inequalities|0N32V0E3XJ]], we get $L \ge 0$.
2. Let $r = (L + 1)/2$. Since $0 \le L < 1$, we have $L < r < 1$ and $r > 0$.
3. Apply the definition of the [[limit|0XKWSSY8P2]] with $\varepsilon = r - L > 0$. There is $N$ with $\abs{a_{n+1}}/\abs{a_n} < r$ for all $n \ge N$.
4. So $\abs{a_{n+1}} < r \abs{a_n}$ for $n \ge N$. By induction on $j$, $\abs{a_{N+j}} \le r^j \abs{a_N}$ for every $j \ge 0$.
5. Let $M \ge N$. By step 4 and the [[sum of a finite geometric series|05APWSX53S]],
$$\sum_{n=N}^{M} \abs{a_n} \le \abs{a_N} \sum_{j=0}^{M-N} r^j = \abs{a_N} \frac{1 - r^{M-N+1}}{1 - r} \le \frac{\abs{a_N}}{1 - r}.$$
The last inequality uses $0 < r < 1$.
6. So for every $M$, $\sum_{n=0}^{M} \abs{a_n} \le \sum_{n=0}^{N-1} \abs{a_n} + \abs{a_N}/(1-r)$. This bound does not depend on $M$.
7. The terms $\abs{a_n}$ are nonnegative and their partial sums are bounded above. By the [[criterion for nonnegative series|0MX1HZBBXJ]], $\sum \abs{a_n}$ converges.
8. So $\sum a_n$ converges [[absolutely|0W3K3QPEWT]].

## Idea

If the ratio of consecutive terms tends to $L < 1$, then eventually each term is at most $r$ times the previous one, for a fixed $r < 1$. So the tail is dominated by a convergent geometric series.