Infinite Sequences and Series BC only
Fifteen topics on sums with no last term. A series converges when its partial sums settle, which is the only definition there is, and terms shrinking to zero is necessary and never sufficient. Most of the unit is tests that decide convergence without computing it: the integral test, whose value is not the sum, the p-series threshold at p > 1 strictly, two comparison pairings that conclude nothing, the alternating test with two conditions, and the ratio test whose value of 1 is no information at all. Then approximation: a Taylor polynomial built on a centre, two error bounds that answer the same question by different means, and an interval of convergence whose endpoints the ratio test cannot see.
AB exam n/aBC exam 15-20%15 topics
Topics
Key forms For every problem in this unit
Converges means
the PARTIAL SUMS Sn settle. Nothing else
nth term test
an NOT → 0 proves DIVERGENCE. That is all
Never
an → 0 does NOT prove convergence
Geometric
a/(1 − r), only when |r| < 1
That a
is the FIRST TERM PRESENT, not the coefficient
p-series
converges when p > 1. STRICTLY. p = 1 diverges
Integral test
positive, continuous, DECREASING. All three
Its value
is NOT the sum. It only decides
Comparison
under CONVERGENT, or over DIVERGENT. Only those
Limit comparison
L finite and POSITIVE gives both verdicts
Alternating
bn DECREASING and bn → 0. Two conditions
Ratio test
L < 1 converges, L > 1 diverges
L = 1
NO INFORMATION. Not divergence
Absolute first
test sum |an| BEFORE sum an
One way only
absolute ⇒ convergent. NEVER the reverse
Alternating bound
|S − Sn| ≤ b(n+1)
Which term
the FIRST OMITTED, not the last used
Why it works
the limit is TRAPPED between consecutive Sn
How many terms
solve b(n+1) < tolerance. Off by one is costly
Lagrange bound
M|x − a|^(n+1) / (n+1)!
All three
order, factorial and power are ALL n+1
That M
bounds the derivative on the WHOLE interval
Not at a point
c is unknown, which is why a bound is needed
Telling them apart
a DERIVATIVE bound means Lagrange
The other way
a numerical series means the ALTERNATING bound
Taylor coefficient
fⁿ(a) / n!, times (x − a)ⁿ
All the same
derivative order, factorial and exponent
The centre
appears TWICE. Maclaurin is just a = 0
General term
substitute n = 0, 1, 2 and CHECK it
Radius
from the ratio test, solving L < 1
Endpoints
give L = 1. The ratio test CANNOT see them
So
substitute each one and test it separately
Four shapes
all occur. No bracket can be assumed
Know these
1/(1−x), eˣ, sin x, cos x
Check a recall
parity, and the value at x = 0
Substituting
put it into the CONDITION too
2x halves
the radius. x² does not change it
Term by term
radius SURVIVES, endpoints may not
Integrating
can GAIN an endpoint. Differentiating can lose one