Stewart Calculus · 8th Edition
Chapter 11: Infinite Sequences & Series
Every key formula from Stewart Calculus Chapter 11, in one searchable page. Click a card to study it — worked examples included.
11.1–11.2 · SEQUENCES & SERIES
{an} converges if lim an = L
Geometric: ∑n=1∞ arn−1 = a1 − r, |r| < 1
Test for Divergence: lim an ≠ 0 ⇒ diverges.
WORKED EXAMPLE
∑n=1∞ (2/3)n = 2/31/3 = 2.
Telescoping series
∑[bn − bn+1] = b1 − lim bn
SN = b1 − bN+1: split into partial fractions first.
WORKED EXAMPLE
∑n=1∞ 1/[n(n+1)] = ∑(1/n − 1/(n+1));
SN = 1 − 1/(N+1) → 1.
SN = 1 − 1/(N+1) → 1.
Integral test
f positive, decreasing, continuous:
∑f(n) and ∫1∞ f agree.
∑f(n) and ∫1∞ f agree.
p-series
∑ 1/np converges ⟺ p > 1
Limit comparison
lim an/bn = L > 0 ⇒ same behavior
Comparison test
0 ≤ an ≤ bn: ∑bn conv. ⇒ ∑an conv.
∑an div. ⇒ ∑bn div. Compare against a p-series or geometric.
WORKED EXAMPLE
∑(n+1)/(n³+2n) vs ∑1/n²:
ratio → 1 > 0; p = 2 > 1 ⇒ converges.
ratio → 1 > 0; p = 2 > 1 ⇒ converges.
11.5–11.6 · ALT., RATIO, ROOT
Alternating series
∑(−1)nbn: bn ↓ 0 ⇒ converges; |Rn| ≤ bn+1.
Ratio / Root
L = lim|an+1/an| or lim ⁿ√|an|
L < 1 abs. conv.; L > 1 div.; L = 1 inconclusive.
WORKED EXAMPLE
∑ n!/nn: ratio → 1/e < 1 ⇒ converges.
Absolute ⇒ convergent
∑|an| converges ⇒ ∑an converges
∑(−1)n/n converges conditionally (harmonic diverges).
WORKED EXAMPLE
∑(2n/(3n+1))n: root test → 2/3 < 1 ⇒
converges absolutely.
converges absolutely.
11.8–11.9 · POWER SERIES
∑ cn(x − a)n
Radius R from the ratio test.
Interval: test endpoints SEPARATELY!
Interval: test endpoints SEPARATELY!
Differentiate/integrate term-by-term (same R).
Radius from ratio/root
R = 1/lim|an+1/an| or 1/lim|an|1/n
R = 0: converges only at x = a; R = ∞: converges for all x.
Substitution trick
e−x² = ∑(−1)nx2n/n!, R = ∞
Plug u = −x² into the eu series — radius unchanged.
WORKED EXAMPLE
∑xn/n²: ratio → |x| < 1 ⇒ R = 1;
x = ±1: ∑1/n² conv. ⇒ interval [−1, 1].
x = ±1: ∑1/n² conv. ⇒ interval [−1, 1].
11.10–11.11 · TAYLOR & MACLAURIN
f(x) = ∑ f(n)(a)/n! · (x − a)n
Famous Maclaurin series
ex = ∑xn/n! · 1/(1−x) = ∑xn (|x|<1)
sin x = ∑(−1)nx2n+1/(2n+1)!
cos x = ∑(−1)nx2n/(2n)! · ln(1+x) = ∑(−1)nxn+1/(n+1)
sin x = ∑(−1)nx2n+1/(2n+1)!
cos x = ∑(−1)nx2n/(2n)! · ln(1+x) = ∑(−1)nxn+1/(n+1)
Taylor’s inequality
|Rn| ≤ M|x − a|n+1/(n+1)!
Binomial series
(1+x)k = ∑(kn)xn, |x| < 1
arctan
tan−1x = ∑(−1)nx2n+1/(2n+1), R = 1
WORKED EXAMPLE
limx→0(ex − 1 − x)/x²:
ex = 1 + x + x²/2 + … ⇒ 1/2.
ex = 1 + x + x²/2 + … ⇒ 1/2.
Maclaurin = Taylor at a = 0.
ln(1+x) = ∑(−1)n−1xn/n, |x| < 1.
WATCH OUT!
Mistakes that cost points
- Divergence test proves DIVERGENCE only
- p ≤ 1 diverges — the harmonic series too!
- Alternating needs BOTH bn → 0 AND decreasing
- Radius ≠ interval — always check endpoints
- The center a shifts the whole series
- Ratio/root with L = 1: switch tests — try comparison
- Binomial series (1+x)k converges only for |x| < 1
- Substituting into ex or sin x keeps the radius (∞ stays ∞)
- Taylor about a ≠ 0: expand in powers of (x − a), not x
- If terms don’t → 0, the divergence test kills the series first
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