Taylor and Maclaurin Series
Derive Taylor series f(x) = ∑(f⁽ⁿ⁾(a)/n!)(x−a)ⁿ and memorize the Maclaurin series for eˣ, sin x, cos x, and 1/(1−x). Construct new series by substitution, differentiation, and integration of known series.
A lesson of AP Calculus BC, in Infinite Sequences and Series, about 10 minutes.
Series and Sequences
By the end of this lesson, you will be able to determine convergence of sequences, apply convergence tests to series, and work with power series and Taylor/Maclaurin series.
Sequences
Sequence: an ordered list of numbers: ${a_1, a_2, a_3, \ldots}$ or ${a_n}$.
Convergence: a sequence ${a_n}$ converges if $\lim_{n \to \infty} a_n = L$ (a finite number). Otherwise, it diverges.
Examples:
- $a_n = \dfrac{1}{n}$: converges to 0
- $a_n = n$: diverges to $\infty$
- $a_n = (-1)^n$: diverges (oscillates)
- $a_n = \dfrac{n}{n+1}$: $\lim_{n\to\infty} = 1$ → converges to 1
Infinite Series
Infinite series: the sum of all terms of a sequence:
$$\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots$$
Partial sums: $S_N = \displaystyle\sum_{n=1}^N a_n$
The series converges if $\lim_{N \to \infty} S_N = L$; the sum is $L$.
Geometric Series
$$\sum_{n=0}^{\infty} ar^n = a + ar + ar^2 + \cdots$$
Converges if $|r| < 1$: sum = $\dfrac{a}{1-r}$
Diverges if $|r| \geq 1$.
Example: $\displaystyle\sum_{n=0}^{\infty} \dfrac{1}{3^n} = \dfrac{1}{1 - 1/3} = \dfrac{3}{2}$
Convergence Tests
Divergence Test (nth-Term Test)
If $\lim_{n \to \infty} a_n \neq 0$ (or doesn't exist), the series diverges.
If $\lim_{n \to \infty} a_n = 0$, the test is inconclusive (the series may still diverge).
Example: $\displaystyle\sum_{n=1}^{\infty} \dfrac{n}{n+1}$: $\lim_{n\to\infty} \dfrac{n}{n+1} = 1 \neq 0$ → diverges.
Integral Test
If $f$ is positive, continuous, and decreasing on $[1, \infty)$, then $\displaystyle\sum_{n=1}^{\infty} f(n)$ and $\displaystyle\int_1^{\infty} f(x), dx$ both converge or both diverge.
$p$-Series: $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^p}$ converges if $p > 1$; diverges if $p \leq 1$.
Special case: harmonic series $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n}$ diverges ($p = 1$).