Andrew Taylor
Alec Seco

Sequence: a function whose domain is a set
of positive integers
◦ Ex. {1,1/2,1/4,1/8,1/16} or (an = ( 1/2 )n-1)

1
2
Either converges or diverges
◦ converge: sequence approaches some number L
lim( an )  L
n 
◦ diverge: sequence doesn’t approach any finite
number
lim(a )   /  or oscillates
n 
n

an )
To test, simply solve lim(
n 
an  e n diverge or converge?
◦ Ex.
lim(e n )  lim( e1n )  0
n
◦ Ex.
n
an  sin(n)
diverge or converge?
lim sin(n)  sin()
n 
converges
diverges due to
oscillation
• Sometimes you might have to use
L’Hopital’s rule
an  (1  )
1 2n
n


Involves multiplying each term by a fixed
term

Generally in the form
ar n 1

n 1
r  ratio
Rule for Geometric Convergence/Divergence
|r| < 1  sequence converges
|r|  1  sequence diverges



Each term involves adding/subtracting a
number to the one before
Ex: {1,3,5,7,9,11}
ALL arithmetic sequences diverge except
when the added term is 0
◦ {0,0,0,0} or {3,3,3,3} (duh.)

Series basically take sequences and add up the
numbers
◦ Again, we look at convergence and divergence



General format for series
k
a
n
n 1
◦ Evaluates the sum of a
sequence from 1st (or 0th) term to any number
Infinite series- what we’ll mostly look at concerning
convergence and divergence; has unlimited number
of terms
Finite series – evaluate from n=1 or 0 to a certain
number
◦ Find the sum of a certain number of terms
◦ Ex: 5
n = 1+2+3+4+5 = 15

n 1

Geometric series are sums of geometric
sequences, similar format

n 1
ar

n 1

Rules of Series convergence/divergence are
the same as sequences
◦ |r| < 1, converge
◦ |r| > or = 1, diverge
**Arithemetic Series**
These never converge
unless they are the zero set
{0,0,0,0} (duh.)



General form:

1

p
n
n 1
Conditions for convergence: p<1
Conditions for divergence: p  1
◦ Special Case: Harmonic Series
 Most common p-series
 diverges

1

n 1 n
***becomes extremely useful with Direct
Comparison Test***


A series where the sign alternates

General form:  (1)n an
n 1

If it meets 2 qualifications, it converges
1.
lim an  0
n 
2.
an  an1

The error will always be smaller than the first
neglected term
◦ Say you found the sum for the first four terms; the
error will always be less than the magnitude of the
fifth term
| RN |
N 1
 a

Ex: Find the error of
(1)n

n
n 1
5

Absolute convergence: just checking if the
absolute value of the series converges
◦ Can be a quick alternative before using Alt. Series
Test
◦ If a series absolutely converges, then it always
converges

General form for a Taylor Series centered at x = a:
f (a) f '(a)( x  a) f ''(a)( x  a)2 f '''(a )( x  a )3
f n (a )( x  a ) n
f ( x  a) 



 ... 
0!
1!
2!
3!
n!


If f(x) is centered at 0, then the power series is
called a Maclaurin Series.
The series can also be written in summation
notation (sigma), from above:


n 0
f n (a)( x  a)n
n!

Given a power series centered at x=a,
exactly one of the following will be true:
◦ The series converges for all x. (-∞,∞)
◦ The series converges for only one x=a.
◦ The series converges on some interval (b,c)… but
in this case you must also check the endpoint
values b and c, assigning closed or open brackets
as necessary.



The following are series that MUST be
memorized:
1. ex - converges for
all
x.
(-∞,∞)
2
3
n
x
x
x
x
e x  1     ... 
1! 2! 3!
n!
2.
1
1 x
- converges on interval (-1,1)
1
 1  x  x 2  x 3  x 4  ...  x n
1 x

3. sin(x) – converges for all x. (-∞,∞)
x3 x5 x 7
(1)n x 2n1
sin( x)  x     ... 
3! 5! 7!
(2n  1)!

4. cos(x) – converges for all x. (-∞,∞)
x 2 x 4 x6
(1)n x 2 n
cos( x)  1     ... 
2! 4! 6!
(2n)!




1. Apply the ratio test to the general term for
the series.
2. Set the resulting expression < 1. (Don’t
forget the absolute value…)
3. Find the interval of convergence.
4. Check endpoints.



Ex.
2
1  x2
But this is really 
1 
2
2 
1

(
x
)

Which is just
with x2 replacing x and
1 by a factor of 2.
then multiplying
1 x



So by applying those two changes to each
term in the known Taylor series for 1
we
can obtain the power series for
1 .x
2
This gives us:
2
1 x
2
And then
(2)1 the
(2)( xanswer:
)  (2)( x 2 ) 2  (2)( x2 )3  ...  (2)( x2 ) n
2  2 x 2  2 x 4  2 x 6  ...  2 x 2 n

A typical question reads like this:
◦ Prove that





f ( x)  P( x)  Q
P(x) is an nth degree Taylor polynomial for a given f(x).
X is some value to be plugged into the polynomial.
Q is the target error.
Showing that the error is less than some value.
General Formula To Use:
R( x)  f ( x)  P( x)  max f

( n 1)
But what does that mean???
(k ) 
xc
n 1
(n  1)!

Here’s what that formula says to do:
◦ 1. Find f ( n1) ( x) , which is the derivative of f(x)
one order higher than the last one you took when
finding P(x). Ex. If P(x) was a third-degree Taylor
polynomial, we are talking about the 4th
derivative.
( n 1)
f
( x)
◦ 2. Find the maximum value
can be
between the value the function is centered at and
your given “x” value.
1
n 1
◦ 3. Multiply this value by (n  1)! and by x  c
 “C” is value f(x) is centered at, so when c=0, its just the
given x value to the n+1 power.