The ratio test; absolute and conditional convergence

Warm-up

Three quick questions from earlier lessons.

Discover

For $\displaystyle\sum \dfrac{2^n}{n!}$ , each term is the previous one times $\dfrac{2}{n + 1}$ . For large $n$ that factor is tiny. Does the series converge?

Yes
No, $2^n$ grows

Key idea

Compare with a geometric series through the ratio of terms.

Ratio test:

L = \displaystyle\lim_{n \to \infty} \left|\dfrac{a_{n+1}}{a_n}\right|

. If

L < 1

,

\sum a_n

converges absolutely; if

L > 1

, it diverges; if

L = 1

, the test is inconclusive.

Worked example

$\displaystyle\sum \dfrac{n^2}{3^n}$ .

\dfrac{a_{n+1}}{a_n} = \dfrac{(n + 1)^2}{n^2} \cdot \dfrac{1}{3}

Divide consecutive terms.

L = \tfrac{1}{3}

(n + 1)²/n² → 1.

Converges absolutely

L < 1.

Your turn, guided

Your turn: $L$ for $\displaystyle\sum \dfrac{5^n}{n}$ .

L = 5

5 · n/(n + 1) → 5 > 1: diverges.

Practice ★

For $a_n = \dfrac{n}{ 8^n}$ , find $L = \displaystyle\lim_{n \to \infty} \left|\dfrac{a_{n+1} }{a_n}\right|$ .
For $a_n = \dfrac{n}{ 4^n}$ , find $L = \displaystyle\lim_{n \to \infty} \left|\dfrac{a_{n+1} }{a_n}\right|$ .
For $a_n = \dfrac{n}{ 5^n}$ , find $L = \displaystyle\lim_{n \to \infty} \left|\dfrac{a_{n+1} }{a_n}\right|$ .

Watch out!

Practice ★★

For $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^{ 2 } }$ , the ratio test gives $L = 1$ . What does it tell you?
1. Nothing: the ratio test is inconclusive when L = 1
2. The series converges
3. The series diverges

How does $\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^n 1 }{n^2}$ converge?
1. Absolutely: ∑ 1/n² converges
2. Only conditionally
3. It diverges

Exit ticket

Three questions to check you have got it.

For $a_n = \dfrac{n}{ 8^n}$ , find $L = \displaystyle\lim_{n \to \infty} \left|\dfrac{a_{n+1} }{a_n}\right|$ .
For $a_n = \dfrac{n}{ 4^n}$ , find $L = \displaystyle\lim_{n \to \infty} \left|\dfrac{a_{n+1} }{a_n}\right|$ .
For $a_n = \dfrac{n}{ 5^n}$ , find $L = \displaystyle\lim_{n \to \infty} \left|\dfrac{a_{n+1} }{a_n}\right|$ .

Summary card

Ratio test

$L = \lim |a_{n+1}/a_n|$ : $L < 1$ converges, $L > 1$ diverges, $L = 1$ unknown.

Watch out: use it for factorials and powers; not for p-series.

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