Calculus II
Spring 2008
Recitation Session #8
(§11.1 and §11.2)
 Summary of important concepts
§11.1
[1] The game of sequences
By definition, an infinite sequence is a function whose domain is the set of positive integers. Can you guess
the function if only first few members of the sequence are given? Of course, trial and error is what people
always adopt, but there are certain rules one needs to know:
(a) The arithmetic sequence: whenever the difference between two consecutive elements in the sequence is
the same
Example: 3, 7, 11, 15, 19, 23,…
The function of this sequence is an  3  (n  1)4  4n  1, n  1, 2,...
The general formula for an arithmetic sequence is an  a1  (n  1)d , n  1, 2,3,... where a1 is the first element
of the sequence, and d is the common difference between consecutive elements.
Keep in mind that you can always check your answer by plugging different n-values to recover the first
few members of the sequence.
(b) The geometric sequence: whenever the ratio between two consecutive elements in the sequence is the same
3 3 3 6 1
3 3 3 3
 
 .
Example: , , , ,...
NOTE using terms 2 and 3 we have
2 6 18 54
18 6 18 3 3
31
The function of this sequence is an   
2 3
n 1

1
n2
3
2
, n  1,2,3,...
NOTE that a1 
3
2
The general formula for a geometric sequence is an  a1r n1 , n  1, 2,3,... where a1 is the first element of the
sequence, and r is the common ratio between consecutive elements.
Keep in mind that you can always check your answer by plugging different n-values to recover the first
few members of the sequence.
(c) Alternating sequence: whenever the sequence changes sign between consecutive elements, use (1) n in
the function.
Example: 1, -4, 7, -10, 13, -16,…
First we recognize that the sequence is arithmetic without the negative sign. Find the function for that
sequence first, an  1  (n  1)3  3n  2, n  1, 2,3,... Then, take care of the alternating signs: an  (1)n 1 (3n  2).
[2] The limit of a sequence: convergent or divergent sequences
Once the function of a sequence is obtained, one can quickly determine whether the limit of the sequence is a finite
number. If the limit of the sequence is not a finite number, we say the sequence diverges, otherwise, it converges.
Note: L’Hopital is in Play.
Example: Does the sequence an 
ln(n  1)
n2
converge?
Let the corresponding continuous function be f ( x) 
lim
ln( x  1)
x2
x 
ln( x  1)
x2
. Take the limit of this function as x  :
1
1
 lim x  1  lim
 0 . Therefore, the sequence converges.
x  2 x
x  2 x( x  1)
Example: lim n n2  1
n
Let the corresponding continuous function be f ( x)  x x 2  1 . Take the limit of this function as x  :
2x
lim ( x  1)
2
1/ x
x
0
 lim e
ln( x2 1)1/ x
x
 lim
x
1
ln( x 2 1)
ex
 lim
x
ln( x 2 1)
e x
ln( x 2 1)
lim
x 
x
e
2
lim x 1
x  1
e
e
lim
2x
x  x 2 1
 e0  1 .
Therefore, the sequence converges.

Get used to the factorial notation and arithmetic:
n!  1  2  3  4...(n  1)n
(n  3)!  1 2  3  4...(n  1)n(n  1)(n  2)(n  3)
6!
What is (n  2)!? What is ?
3!
(n  1)!
1  2  3  4...(n  2)(n  1)
1
Let’s simplify


(n  2)! 1  2  3  4...(n  2)(n  1)n(n  1)(n  2) n(n  1)(n  2)
A useful result that can be used as a formula:
xn
n!
 0, and lim n  ,for any x. We explained this in
n n!
n x
lim
class.
3n 20n 1
3n 20n 201
(3  20)n
60n
 lim
 201 lim
 201 lim
 0.
n
n
n
n n !
n!
n!
n!
Example: lim
nn
n!
  and lim n  0 . We explained this in class.
n n!
n n
Note: lim
§11.2
[1] The notations of series
N
Partial sum: S N   an  a1  a2 ...  aN (with finite number of terms)
n 1

Infinite series:  ak  a1  a2 ...  aN  aN 1  ... (with infinite number of terms, you may use any letter for the
k 1
running index)

N
Partial sum and the infinite series:  ak  lim S N  lim  an  L . If L is finite, we say the series converges;
N 
k 1
N  n 1
otherwise, the series diverges. Obviously, the following series diverges:

 2 because the partial sum
n 1
N
S N   2  2  2  2  ...  2  2 N , and lim S N  lim 2 N   .
n 1
N
N 
N 
[2] Serious Series (say this ten times in a row, have fun)
(a) Geometric series:

a  ar  ar 2  ...  ar N 1  ...   ar n1 : Obviously, geometric series is obtained from adding together the
n 1
members of an infinite geometric sequence. Do you remember how to verify a given sequence is a
geometric sequence? That is the same approach used to verify whether a given series is a geometric
series. A more general geometric series looks like:
ar k  ar k 1  ar k  2  ... 

 ar n
nk
A formula for the sum of a partial sum of a geometric series:
SN 
N
 ar n 
nk
first term  last term  r
1 r
1 1 1 1 1


1 1
1
1
9841
9 310 3 9 311
Example:    ...  10 
.


1
2
9 27 81
59049
3
1
3
3
Important result: the geometric series converges if r <1; the series diverges if r  1 .
An important formula (slightly different from the one on your book): if r  1 , the sum of the

convergent geometric series is:  ar n 
nk
1
1 1
1
Example:    ...  9 
1
9 27 81
1
3
first term
.
1 r
1
9  1.
2 6
3
A classical example: rewrite the repeated decimals into a rational number
3.121121121...  3 
121
121
121
121/1000
121 3118


 ...  3 
 3

.
1
1000 10002 10003
999
999
1
1000
In other words, the geometric series begins in the example above with the 2nd term, after the 3. Then add 3.

(b) Telescopic series: A telescopic series can always be written as   an  an1  , as the series is written out,
nk
most of the terms in the partial sum are cancelled. It is a fun problem to solve.

1
A
B
 
n(n  2) n n  2
n 1
First, find the partial sum of the telescopic series:
1
 n(n  2)
 note  use partial fractions:
N 1/ 2
1
1/ 2  1 N  1
1 

 

  

n  2  2 n 1  n n  2 
n 1 n( n  2) n 1  n
N
SN  
 1 1   1 1   1 1   1 1   1 1   1 1 

 1  3    2  4    3  5    4  6    5  7    6  8   ... 
1 
 
 
 
 
 




2  1
1   1
1  1
1  1
1 







  N  3 N  1   N  2 N   N  1 N  1   N N  2  

1 1 1
1  1 3 1
1 
1   
   

2  2 N N  2 2 2 N N  2

1
SN
 n(n  2)  Nlim

n 1
1 3 1
1  3
  

N  2  2
N N  2 4
 lim
[3] Test for Divergence

Important: If lim an  0 , then the series  an diverges.
n 
 Test for divergence
n 1

Example:  cos n diverges because lim cos n  lim (1) n does not exist
n 
n 1

 sin n diverges because
n 1

lim sin n does not exist
n
1
 cos( n ) diverges because
n 1
n
1
lim cos( )  cos(0)  1  0
n
n
Strategy: Always try Test for Divergence first whenever testing a series.
Common Error: Many students make this mistake: If lim an  0 , it does not necessarily imply that the series
n 
converges!!! Keep in mind, lim an  0 does not tell you anything about the series!!!
n 
1
 0 , however, the series
n n
Example: It is obvious that lim

1
 n (harmonic series) still diverges.
n 1
Homework:
§11.1: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 60, 61, 62, 65.
§11.2: 1, 3, 5(telescopic series), 7, 9, 11, 13, 19, 21, 22, 24, 25(geometric series), 27, 29(geometric series), 31, 33, 35,
37 (telescopic series), 39, 51, 53, 56.
Calculus II
Recitation Session #8
(§11.1 and §11.2)
NAME: _______________________________
(25-points total)
(1-3 pts) Find a formula for the sequence 1, 4,7, 10,13, 16,...
(2-6pts) Find the limit of the sequences, use L’Hopital’s rule whenever needed:
(A-3pts) an
1
 (ln n) n
(B-3pts) bn 
(3-6pts) Find the sum of the geometric series
 3(1) n  22 n 

5n

n2 



2n 1100n
n!
(4-8pts) Find the sum of the telescopic series (show brackets and details of work)

2
 (n  1)(n  1)
n2

n
converge or diverge? Why?
n 1 2n  1
(5-2pts) Does the series 