Series Flashcards
(20 cards)
Geometric Series
E ar ^n Sum = a / (1-r) Converges: |r|<1 (limit = 0) Diverges: |r|>=1 (limit approaches infinity) MUST INCLUDE |r| < 1!
*Geometric Series (r, ar, when n = 1)
- whatever’s raised to the Nth degree is r (when series starts at 0)
- ar = 2nd term
- r = 2nd term / 1st term
a = …, ar = .., ar = ar (divide to get r)
Reindexing a Geometric Series
- If given summation of starting at n = 1, E #^n+1 / #^n-1 …
- If given summation of starting at n = 1, E (-1)^n-1 # / #^n
- so the true value of ‘a’ can be seen
Raise: start @n = 0, E #^(n+1)+1 / #^(n+1)-1
[ simplify, state a,r, |r| > 1 ]
n = # + 1, subtract 1 from starting n
Raise: start at n = 0, E (1-)^(n+1)-1 #/ #^(n+1)
[ simplify, state a,r, |r| > 1 ]
n = # + 1, subtract 1 from starting n
- Lower:
n = 1 - h, add h to an
Nth Term Test DIVERGENCE
if lim of an (n approaches infinity) DOES NOT EQUAL ZERO, it diverges
*lim = 0 means NOTHING (inconclusive)
*Telescoping Series Method
- has both a positive and negative term
- Sk = the series at (n=1) + (n=2) + (n=3) +…
… + (n=k-2) + (n=k-1) + (n=k) - state sk after canceling out first and last
- take the limit
- MAY HAVE TO USE PARTIAL FRACTIONS
Integral Test
- compare to
- integral from n to infinity
- replace n’s with x’s (dx)
- check positive ( > 0), decreasing (f’(x) < 0 OR an > an+1), solve integral (infinity = diverges, finite = converges)
P-Series
- Summation of 1 / n ^p
- p = 1, harmonic series (1/n DIVERGENT)
- p > 1, converges p < 1, series diverges
Direct Comparison Test
- given an choose bn, COMPARE LEADING TERMS
- state what bn does and how you know
- if bn converges, prove an < bn
- if bn diverges, prove an > bn
anything else = inconclusive
Limit comparison
- given an choose bn, COMPARE LEADING TERMS
- prove positive
- lim (n-> infinity) an / bn = # > 0, both converge or diverge
- lim (n-> infinity) an / bn = 0, AND bn CONVERGE, then an converges
- lim (n-> infinity) an / bn -> infinity, AND bn DIVERGE, then and diverges
- else: lim (n->infinity) bn / an = conclusive answer
Alternating Series Test (AST)
(-1)^n or (-1)^n+1
- must exist to use AST, but is IGNORED when testing
- positive, decreasing (an > an+1), lim as n -> infinity = 0
Absolute Convergence
- compare to series with abs values, if converges, converges
- root / ratio test if lim < 1 converges absolutely
- power series, get limit into |x-a| < r form
- -1 < |x-a| < 1 int of abs conv
Ratio Test
- must prove positive or use abs values
- lim of an+1 / an as n -> infinity
- < 1, converges absolutely
- > 1 or -> infinity, diverges
- = 1, INCONCLUSIVE (usually with p-series)
Root Test
- must prove + or use abs values
- lim of the nth root of an as n -> infinity
- < 1, converges absoulutely
- > 1 or -> infinity, diverges
- =1, INCONCLUSIVE (usually with p-series)
Power Series
- has a x in it
Testing Power Series for Absolute Convergence
- use Root / Ratio test to find interval of absolute convergence
- |x-a| < R
Find the power series representation and radius of convergence
f(x) = integral E
-> f’(x) -> E ->
-> integral f(x)dx -> E -> f(x) = d/dx E
get into i / a - r form, write converges for |r|<1, simplify to get R
Interval of Convergence
- solve limit
- | | < 1
- put x term in the middle of -1 1 inequality and simplify
- -> interval of absolute convergence
- test endpoints (conditional convergence) together = int. of conv.
Taylor Polynomial vs Maclaurin + formula
Taylor a = given, Maclaurin a = 0 always
- derivatives, evaluate
- f(a) + f’(a)(1/1!)(x-a) + f’‘(a)(1/2!)(x-2)^2 + f’’‘(a)(1/3!)(x-a)^3 + …
- order = p(0), p(1), p(2)…
Binomial series (first 4 terms)
m = exponent of quantity
if x has coefficient x = “_x”
1 + mx + m(m-1)(1/2!)x^2 + m(m-1)(m-2)(1/3!)x^3 + …
Built off of (1 + x)^m
Odd Numbers, Even Numbers
Odd: 2n + 1, Even: 2n
