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Series and Convergence

Convergence tests and Taylor series.

An infinite series adds infinitely many terms, $\sum_{n=1}^{\infty} a_n$ . It converges if its partial sums approach a finite limit. The geometric series $\sum ar^{n}$ is the model case: it converges to $\frac{a}{1 - r}$ exactly when $|r| < 1$ , which is why $1 + \frac{1}{2} + \frac{1}{4} + \cdots = 2$ .

Convergence tests decide whether other series settle down. The terms must shrink to zero, but that alone is not enough: the harmonic series $\sum \frac{1}{n}$ diverges. Power series and Taylor series then represent functions such as $e^x$ and $\sin x$ as infinite polynomials, which is how calculators evaluate them.

Key ideas

$n$ th-term test: if $a_n \not\to 0$ , the series diverges.
$p$ -series $\sum \frac{1}{n^p}$ converges exactly when $p > 1$ .
Ratio test: $\lim \left|\frac{a_{n+1}}{a_n}\right| < 1$ means absolute convergence.
Taylor series: $e^x = \sum \frac{x^n}{n!}$ , valid for all $x$ .

Worked example

Find the sum of $6 + 2 + \frac{2}{3} + \frac{2}{9} + \cdots$ .

First term $a = 6$ ; ratio $r = \frac{2}{6} = \frac{1}{3}$ .
$|r| < 1$ , so the series converges.
Sum $= \frac{6}{1 - \frac{1}{3}} = \frac{6}{\frac{2}{3}} = 9$ .

Answer: $9$

Common mistakes

Concluding convergence just because the terms go to zero.
Using the geometric formula when $|r| \ge 1$ .
Starting the sum at the wrong index and getting the first term wrong.

Practice questions

Try each one, then open the answer. Numerical answers on this page are verified by an automated calculation check.

1.Sum $\sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^n$ .
Show answer
$2$
2.Sum $5 - \frac{5}{4} + \frac{5}{16} - \cdots$ .
Show answer
$4$
3.Use four terms of $\sum \frac{x^n}{n!}$ to estimate $e^{0.1}$ to four decimals.
Show answer
$1.1052$

Series and Convergence

Key ideas

Worked example

Common mistakes

Practice questions

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Calculators

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