Calc Week 5 Midterm Study Flashcards
Evaluating the limit of a sequence
- by calculation and algebraic limit theorem
-multiply numerator and denominator by 1/ biggest term of all
-simplify
-take the limit - thinking of f(x)
-set a sub n to an equation
-set f(x) = a sub n with x plugged in
-take limit of f(x) usually involves L’H
-answer will be lim n-> inf of a sub n - multiplying conjugates
-an = equation of the sequence
-multiply by a fraction (numerator and denominator are that same a sub n but opposite sign)
-simplify
-take limit n -> inf
writing the partial sums of a series
the next s sub n includes the previous s sub n plus the current s sub n
the connection between partial sums and infinite series is
lim n-> inf s sub n =
the sum of the series a sub n
geometric series test
-when to use
-definition
-how to use
-explanation
for a geometric series
r^n
1.the nth partial sum is s sub n = 1-r^n+1/1-r
2. the series converges if and only if |r| < 1 to 1/1-r
- make sure it is a geometric series
2.look at r
3.use test
since |r| < 1, the series converges to 1/1-r by the geometric series test
since |r|>=1, the series diverges by the geometric series test
p-series test
-when to use
-definition
-how to use
-explanation
for a p-series
1/n^p for 1/(an+b)^p, p>0
a general p-series converges if and only p>1
1.make sure it is a p-series
2. look at p
3. use test
since p>1, the series converges by the p-series test
since p<=1, the series diverges by the p-series test
direct comparison test
-when to use
-definition
-how to use
-explanation
for when you can find another series that is larger or smaller that you know converges or diverges
suppose {a sub n} and {b sub n} satisfy o<=a sub n<= b sub n for all n then
1.if series b sub n converges, then series a sub n converges
2.if series a sub n diverges, then series b sub n diverges
- find another series you know that converges or diverges
- figure out what the new series does and if it larger or smaller than original
- use test
since series a sub n or b sub n converges or diverges, the direct comparison test implies series a sub n or b sub n converge or diverge
limit comparison test
-when to use
-definition
-how to use
-explanation
for when you can find a similar/comparable series that you know converges or diverges
let {a sub n} and {b sub n} be sequences such that a sub n>= 0 and b sub n>0 for all n then
1.if lim n-> inf a sub n/b sub n =L, with 0<L<inf the a sub n series converges if and only if the series b sub n converges (works for divergence too)
2.if lim n-> inf a sub n/b sub n = 0 and series b sub n < inf then series a sub n < inf
3. if lim n-> inf a sub n/b sub n= +inf and series b sub n = +inf then series a sub n = +inf
1.find b sub n similar to a sub n
2.take the lim of a sub n /b sub n to see if comparable
3.use test
since L is finite and positive, series a sub n and b sub n are comparable, therefore, the limit comparison test implies since series b sub n converges, so does a sub n
ratio test
-when to use
-definition
-how to use
-explanation
when you see factorials
let {a sub n} be a positive sequence with lim n-> inf a sub n+1/a sub n = L then
1.if L<1, then series a sub n converges
2.if L>1, then series a sub n diverges
3.if L=1, the test is inconclusive
- find a sub n
2.take lim n-> of a sub n+1/a sub n
—plug in n+1 and n into a sub n - use test
since L = # >/< 1, by the ration test, the series converges/diverges
alternating series test
-when to use
-definition
-how to use
-explanation
for an alternating series
(-1)^n (a sub n)
(-1)^n+1(a sub n)
–a sub n is positive
let {a sub n} be a positive, monotonically decreasing sequence such that lim n-> inf a sub n = 0 then the series converges
1.find a sub n
2.is a sub n monotonically decreasing
3.is the lim n-> inf of a sub n =0
4.use test
since a sub n is monotonically decreasing and lim n-> inf = 0, then the series converges by the alternating series test
n-term test
-when to use
-definition
-how to use
-explanation
first thing to check to see if automatically diverge
consider the series of a sub n
if lim n-> inf a sub n != 0, the series of a sub n diverges
- find a sub n
- take the lim n-> inf a sub n
since the lim n-> inf of a sub n does not equal 0, the series converges by the n-th term test
other things to know
lim n-> of
(1+a/n)^n
e^a
other things to know
lim x->0 of sinx/x
1
other things to know
absolute value theorem
for a sequence
let {a sub n} be a sequence
if lim->inf of |a sub n |=0 then lim n->inf of a sub n =0
other things to know
power series expansion of e^x
series from n=1 to inf of
x^n/n!
=
e^x
factorial beats exponential
other things to know
absolute convergence theorem
suppose series of |a sub n| converges, then
1.series a sub n converges
2.the value of series of a sub n does not depend on the order of the terms
other things to know
definition of converging absolutely or conditionally
must know if the series even converges
must converge for either one to be true
let series of a sub n be any series
1. if series |a sub n| converges, the series converges absolutely
2.if series a sub n converges, but series |a sub n| = +inf then the series converges conditionally
review
7 tests and when to use, definition, steps and explanation
evaluating a sequence
writing partial sums
the connection
the limit/or sum of 3 (fact) series
3 theorems
——-other things to know—–
algebraic limit theorem
convergence theorem
monotone convergence theorem
integral test
showing comparability (in limit comparison test)
root test
