Math 2J - Series Summary
1 . The series converges if p > 1 and diverges np n=1 if p 1. When p = 1, this is known as the Harmonic series. 1 Example: converges since it is a p-series with p = 2 > 1. n2 n=1
p-series: A series of the form
Geometries Series: A series of the form
i=0
Arn . A can be thought of as
the rst term of the series, and r as the scaling factor. The series diverges if A . |r| 1. If |r| < 1 then the series converges to 1r 1 1/5 Example: converges to 1(1/5) = 1 since the rst term, A, is 1/5 6 (5)n n=1 and r = 1/5.
Telescoping Series: Any series which can be factored and broken up (usually using partial fraction decomposition) so that all of the middle terms of the partial summations cancel out. 1 1 1 = . If you expand sn and simplify, sn = Example: n(n + 1) n=1 n n + 1 n=1 1 1 n+1 , and so sn 1 (i.e. the series converges to 1). n ln . If you expand sn and simplify, sn = ln(n + 1), and Example: n+1 n=1 so sn (i.e. the series diverges to )
Absolutely Convergent Series: A series
n=1
an is absolutely convergent
if
n=1
|an | converges.
Example: verges.
n=1
(1)n converges absolutely because n2
n=1
(1)n n2
=
n=1
1 conn2
Conditionally Convergent Series: A series
n=1
an is conditionally con-
vergent if
n=1
|an | diverges, but
n=1
an converges.
(1)n converges while Example: n n=1 verges conditionally.
n=1
1 diverges, hence the rst series conn
Test for Divergence: If the sequence an does not converge to 0, then the
series
n=1
an diverges.
Example:
n=1
n n 1 diverges because lim = = 0. n 2n + 5 2n + 5 2
1 1 diverges, even though lim = 0. That is, the converse of this n n n n=1 test is NOT true. an may converge to 0, but the series can still diverge. Example:
Integral Test: If there is a continuous, decreasing, positive function f (x)
so that f (n) = an , then the series
n=1
an converges if and only if the integral
f (x)dx converges.
1
Notes: A simple way to show that f (x) is decreasing is to show that the f (x) < 0. 1 Example: diverges. To use the intergral test you need to show that n ln(n) n=2 1 1 + ln(x) f (x) = is always positive (it is), and that its decreasing (f (x) = 2 < x ln(x) x ln(x)2 1 0). Then its a simple matter of taking dx = . x ln(x) 2
Comparison Test: If an and bn are two sequences, and an bn 0, then:
(a) If
n=1
an converges, then
n=1
bn also converges. an also diverges.
(b) If
n=1
bn diverges, then
n=1
Notes: To show that a series diverges, the goal is to nd a sequence whose terms are all smaller, but whose series diverges. To show that a series converges, the goal is to nd a sequence whose terms are all larger, but whose series converges.
Example:
n=1
n n n 1 diverges because 2 > 2 = whose series diverges. n2 1 n 1 n n n2 1 1 1 converges because 2 < 2 whose series diverges. +1 n +1 n
Example:
n=1
Limit Comparison Test: If an and bn are two sequences of postive numbers, and if lim an exists and is nonzero, then n bn
an if and only if
n=1 n=1
bn con-
verges. Notes: This test say that if you have two positive sequences and their ratio has a positive (but nite) limit, then their series behave the same way. 1 Example: converges because we can compare it with the similar 3n2 + 2n + 3 n=1 1 n2 1 but simpler series 2 : lim = 1/3, and since 3 a positive, n n 3n2 + 2n + 3 1 nonzero, nite number, and since the series we compared it with converges, the original series converges too.
Alternating Series Test: If an > 0, and an+1 < an , and n an = 0, then lim
(1)n an converges. Notes: Remember that sin(n + /2) = cos(n) = (1)n are common ways of representing alternating series. Notes: This test says that if you have an alternating series whose sequence terms converge to 0, then it suces to check that the sequence terms are decreasing to show that it converges. As with the integral test, often the easiest way to test for a decreasing sequence is treat it as a function of x and show that the derivative is negative.
n=1
Example:
n=1
(1)n sin(1/n) converges because in quadrant 1 (where 1/n eventu-
ally always lies) sin is decreasing, and sin(1/n) sin(0) = 0 as n .
Root Test: Let n lim
an+1 = c (if the limit does not exist, this test cannot an
be used). If c > 1, then the series
n=1
an diverges. If c < 1, then
n=1
an converges
absolutely. If c = 1, this test does not provide any information. Notes: This test is primarily useful when its very easy to take nth roots, e.g. when the entire expression is raised to the nth power.
n Example: (1)n ( 21)n converges absolutely since lim n n=1 n lim ( 2 1) = 0 < 1. n
n (1)n ( 2 1)n =
Ratio Test: Let n lim
|an | = c (if the limit does not exist, this test cannot
be used). If c > 1, then the series
n=1
an diverges. If c < 1, then
n=1
an converges
absolutely. If c = 1, this test does not provide any information. Notes: This test is primarily useful when taking successive ratios will result in a lot of stu cancelling out, e.g. when factorials are involved. This test is also used when determining the radius of convergence of a power series, Maclaurin series, or Taylor series. (n + 1)2 (2n + 1)! n2 Example: converges absolutely since lim = n (2n + 3)! (2n + 1)! n2 n=1 (n + 1)2 =0<1 lim n n2 (2n + 3)(2n + 2)
