MAT 146
Take-Home Assignment #3
In Take-Home Assignment #2, we defined and explored sequences, which are ordered lists of terms. For TakeHome Assignment #3, we consider a closely related mathematics concept called a series.
A series is the sum of the terms in a sequence. Here is a finite sequence:
v1 = 1, v2 = 4, v3 = 9, v4 = 16, v5 = 25
Here is the series associated with this sequence:
v1 + v2 + v3 + v4 + v5 = 1+ 4 + 9 +16 + 25 = 55
Please use this example to never forget how sequences and series differ: A sequence is a set of terms, finite or
infinite in number, possessing some order based on how the terms are listed. The series associated with a
sequence is precisely one value, where the value is the sum of the terms in the corresponding sequence,
provided that such a value exists. Reread this paragraph several times!
(A) For each finite sequence, calculate the associated series. Show the terms and the sum for each series. Show
the exact value and a decimal approximation to the nearest ten-thousandth of a unit. (5 pts)
Finite Sequence
Associated Series: Show Terms and Sum
vn = n 2 , 1 ! n ! 5
1+ 4 + 9 +16 + 25 = 55
1
bn = , 1 ! n ! 4
n
dn =
2
, 1! n ! 5
n ( n +1)
! 1 $n
gn = 9 # & , 1 ' n ' 6
" 10 %
1
hn = , 0 ! n ! 5 (0! = 1)
n!
n+1
kn =
4 (!1)
, 1" n " 6
2n !1
Thus far, we have considered finite sequences and their associated series. Most of our attention, however, will
focus on series associated with infinite sequences. We call these infinite series, and our interest in them is twofold:
• For a given infinite sequence, can we determine whether its associated infinite series—its sum—takes
on a finite value?
• If an infinite sequence does indeed have a finite sum, what is that sum? That is, what is the value of the
infinite series?
Addressing these two questions is the focus of much of Chapter 11. It turns out that the second question is
generally much tougher to answer. Stay tuned!
To help determine whether an infinite sequence does indeed have a finite sum, we can explore one particular
sequence associated with the original infinite sequence. That associated sequence is called the sequence of
partial sums. Here’s an example.
n
!1$
Suppose an infinite sequence is represented by the general term an = # & , for positive integers n ≥ 1. The first
"2%
five terms of this sequence are
1
1
1
1
1
.
a1 = , a2 = , a3 = , a4 = , a5 =
2
4
8
16
32
We will represent the associated sequence of partial sums using Sn, defined as follows:
A sequence of partial sums Sn represents the sum of the first n terms of a sequence. For a sequence an we have
S1 = a1, S2 = a1 + a2 , S3 = a1 + a2 + a3, …, Sn = a1 + a2 + a3 +!+ an
These partial sums form a sequence, because they represent an ordered list of terms: S1, S2 , S3,…, Sn .
n
!1$
1
1
1
1
1
For the sequence an = # & described previously, where a1 = , a2 = , a3 = , a4 = , a5 = , the associated
"2%
2
4
8
16
32
sequence of partial sums is
S1
= a1
S2
= a1 + a2
S3
= a1 + a2 + a3
Sn
= a1 + a2 + a3 +!+ an
1
2
1 1
= +
2 4
1 1 1
= + +
2 4 8
1 1 1
1
= + + +!+ n
2 4 8
2
=
Completing the addition, we have this sequence of partial sums:
1
3
7
15
31
S1 = , S2 = , S3 = , S4 = , S5 =
2
4
8
16
32
(B) For each infinite sequence, determine the first five terms of that sequence’s partial sums. That is, determine
S1 through S5, similar to the examples above. Show exact values and decimal approximations to the nearest tenthousandth of a unit. (5 pts)
Infinite Sequence
vn = n 2 ,
n !1
n
!1$
an = # & , n ' 1
"2%
1
bn = , n ! 1
n
2
dn =
, n !1
n ( n +1)
Sequence of Partial Sums Sn, for n = 1,2,…,5
S1 = 1 S2 = 5 S3 = 14 S4 = 30 S5 = 55
1
3
7
15
31
S1 = , S2 = , S3 = , S4 = , S5 =
2
4
8
16
32
1
3
S1 = = 1 S2 = = 1.5 S3 = ? S4 = ? S5 = ?
1
2
! 1 $n
gn = 9 # & , n ' 1
" 10 %
1
hn = , n ! 0
n!
n+1
kn =
4 (!1)
, n "1
2n !1
A sequence of partial sums, Sn, provides evidence to help us determine whether the series has a finite sum. We
use S to represent the value of the series. If the sum exists, S is a finite real number. Please note the distinction
between the series S and the sequence of partial sums Sn. The series S is a single value, if it exists. The partial
sums Sn form a sequence. Reread this paragraph several times!
How do we use the sequence of sums, Sn, to determine S? By looking at the sequence of partial sums, we may
be able to detect a pattern that reveals whether S exists. Return to the example above.
n
!1$
1
3
7
15
31
For an = # & , we computed S1 through S5: S1 = , S2 = , S3 = , S4 = , S5 =
. Is there a pattern revealed in
"2%
2
4
8
16
32
these partial sums that allows us to write the general term, Sn, of these partial sums? We can rewrite the first
five partial sums to help identify one such pattern.
S1 =
1 2 !1 21 !1
3 4 !1 2 2 !1
7 8 !1 23 !1
15 16 !1 2 4 !1
31 32 !1 2 5 !1
=
= 1 , S2 = =
= 2 , S3 = =
= 3 , S4 = =
= 4 , S5 =
=
= 5
2
2
4
4
8
8
16
16
32
32
2
2
2
2
2
There is a connection here between the index value (the subscript that indicates the location of the term in the
infinite sequence) and the term itself. This leads to a conjecture for a compact representation for the nth partial
sum:
2 n !1
Sn = n
2
(C) For each infinite sequence, create the associated sequence of partial sums for at least n ! {1, 2, 3, 4, 5} . For
each sequence of partial sums, explore the patterns revealed and write an expression for Sn, the general term in
that sequence of partial sums. Your expression should be in terms of n. (5 pts)
Infinite Sequence
General Term for the Sequence of Partial
Sums Sn
vn = n 2 , n ! 1
Sn =
n ( 2n +1) ( n +1)
6
n
!1$
an = # & , n ' 1
"2%
Sn =
2 n !1
2n
! 1 $n
gn = 9 # & , n ' 1
" 10 %
pn =
3n !1
, n "1
3n
n'1
!1$
rn = 2 # & , n ( 1
" 3%
tn =
4 n+1
, n !1
5n
With an accurate representation for the nth partial sum, Sn, we can use limits to determine whether S is a finite
value, because, when S exists, we know that
lim Sn = S
n!"
n
!1$
2 n !1
For the infinite sequence we have followed above, an = # & , we conjectured that Sn = n . If this is true, we
"2%
2
can use it to determine S, the value of the series associated with the original sequence. Here, we have
$ 2n 1 '
$
2 n #1
1'
S = lim Sn = lim n = lim && n # n )) = lim &1# n ) = 1# 0 = 1 = S
n!"
n!" 2
n!" % 2
2 ( n!" % 2 (
n
!1$
1 1 1
1
This shows that the terms in the infinite series an = # & sum to 1. That is, that S = + + +!+ n +! = 1 .
"2%
2 4 8
2
When a finite sum S exists, we say that the infinite series converges.
If the limit of the partial sums does not exist, then we know the series S associated with the original sequence
3n!1 !1
cannot be expressed as a finite real value. For example, consider the sequence cn =
. We have
3n
2
8
26
80
c1 = 0, c2 = , c3 = , c4 = , c5 =
9
27
81
243
and therefore
S1 = 0, S2 = 0 +
2 2
2 8 14
2 8 26 68
2 8 26 80 284
= , S3 = 0 + + = , S4 = 0 + + +
= , S5 = 0 + + + +
=
9 9
9 27 27
9 27 81 81
9 27 81 243 243
It seems the denominators are all powers of 3, but any pattern in the numerators may be difficult to detect.
Examine decimal approximations for these partial sums:
S1 = 0, S2 =
2
14
68
284
1094
! 0.222222, S3 =
! 0.518519, S4 =
! 0.839506, S5 =
! 1.16872, S6 =
! 1.50069
9
27
81
243
729
As n grows larger, it seems that each sum is increased by approximately 1/3 of a unit compared to the previous
3n!1 !1
sum. This makes sense when you consider cn =
, because lim cn = 1 . Each new term in the sequence cn
n
3
n!"
3
adds approximately 1/3 to each new partial sum in the sequence Sn. This is significant because the partial sums,
Sn, just continue to grow larger and larger. There is no upper bound for these partial sums and therefore the
series associated with the sequence cn does not take on a finite value. When this happens, we say that the
infinite series diverges.
(D) Return to the partial sums you explored in exercise (C) as well as to the original sequence from which you
created each sequence of partial sums. Use the original sequence and the associated partial sums to determine
whether S exists for each infinite sequence. Describe and explain the basis for each decision. (10 pts)
Infinite Sequence
vn = n 2 , n ! 1
n
!1$
an = # & , n ' 1
"2%
3n!1 !1
cn =
, n "1
3n
Does S exist? Explain, including “S converges” or
“S diverges.” If S converges, determine the value of S
and explain how you determined it.
For Sn =
n ( 2n +1) ( n +1)
, S = lim Sn = " . Because the
n!"
6
limit of the partial sums, as n grows without bound, is
not a finite value, the series S diverges.
$ 2n 1 '
.
$
2 n #1
1'
S = lim S = lim
= lim & # ) = lim 1#
= 1# 0 = 1 = S
n!"
n
n!"
2n
&
n!" % 2 n
)
2n (
&
n!" %
)
2n (
Because the limit of the partial sums, as n grows
without bound, is a finite value, the series S converges
to 1.
Each new term in the sequence cn adds approximately
1
/3 to each new partial sum in the sequence Sn. Thus,
the partial sums, Sn, continue to grow larger and larger
without bound. Because there is no upper bound for
these partial sums, the series associated with the
sequence cn does not take on a finite value. The series
S diverges.
n
!1$
gn = 9 # & , n ' 1
" 10 %
3n !1
, n "1
3n
n'1
!1$
rn = 2 # & , n ( 1
" 3%
pn =
tn =
4 n+1
, n !1
5n
Resources
http://www.teacherweb.com/TX/FriendswoodHighSchool/MrsGillespie/Sequences-on-the-ti-89.pdf (PDF file: TI-89 and sequences)
http://education.ti.com/html/t3_free_courses/calculus89_online/mod21/mod21_lesson1.html (web site: sequences and series, TI-89 connections)
http://www.youtube.com/watch?v=NpqSFfeXbqU (video: summation on the TI-89)
http://www.youtube.com/watch?v=fEP8XT8hiBg&list=PLAA6688F60255BEFD&index=59 (long video: sequences/series)
Guidelines for Completion of Take-Home Assignment #3

The assignment is designed to be completed in 60 to 90 minutes.

You are encouraged to work with as many as three other people to complete the assignment and to submit one set of responses
for your group. You may work with students in any of my three sections of Calc II
(Section 1: 9 am, Section 5: 12 noon; Section 6: 1 pm).

Record all of your responses on this submission page. You therefore need to submit just one piece of paper.

Point values are indicated for each response. Your responses will be evaluated for accuracy, clarity, and completeness. Your
score will be included in the Quiz category of your semester grades.

Email me any questions. The completed responses are due in my hands when class begins on Monday, October 28, 2013.
Name (1) _______________________________________________________________________ Class Section __________________
Name (2) _______________________________________________________________________ Class Section __________________
Name (3) _______________________________________________________________________ Class Section __________________
Name (4) _______________________________________________________________________ Class Section __________________
Score ____________________ /25
(A) (5 pts)
Finite Sequence
Associated Series: Show Terms and Sum
vn = n 2 , 1 ! n ! 5
1+ 4 + 9 +16 + 25 = 55
1
bn = , 1 ! n ! 4
n
dn =
2
, 1! n ! 5
n ( n +1)
! 1 $n
gn = 9 # & , 1 ' n ' 6
" 10 %
hn =
1
, 0!n!5
n!
kn =
4 (!1)
, 1" n " 6
2n !1
n+1
(B) (5 pts)
Infinite Sequence
Sequence of Partial Sums Sn, for n = 1,2,…,5
vn = n 2 , n ! 1
S1 = 1 S2 = 5 S3 = 14 S4 = 30 S5 = 55
! 1 $n
an = # & , n ' 1
"2%
S1 =
1
bn = , n ! 1
n
S1 =
S2 =
S3 =
S4 =
S5 =
2
, n !1
n ( n +1)
S1 =
S2 =
S3 =
S4 =
S5 =
! 1 $n
gn = 9 # & , n ' 1
" 10 %
S1 =
S2 =
S3 =
S4 =
S5 =
S1 =
S2 =
S3 =
S4 =
S5 =
S1 =
S2 =
S3 =
S4 =
S5 =
dn =
hn =
1
,
n!
n!0
n+1
kn =
4 (!1)
, n "1
2n !1
1
3
7
15
31
= 0.5 S2 = = 0.75 S3 = = 0.875 S4 = = 0.9375 S5 =
= 0.96875
2
4
8
16
32
(C) (5 pts)
Infinite Sequence
vn = n 2 ,
n !1
n
!1$
an = # & , n ' 1
"2%
General Term for the Sequence of Partial Sums Sn
Sn =
n ( 2n +1) ( n +1)
6
2 n !1
Sn = n
2
! 1 $n
gn = 9 # & , n ' 1
" 10 %
3n !1
, n "1
3n
n'1
!1$
rn = 2 # & , n ( 1
" 3%
pn =
tn =
4 n+1
, n !1
5n
(D) (10 pts)
Does S exist? Write “S converges” or write “S diverges.”
If S converges, determine the value of S.
Infinite Sequence
vn = n 2 , n ! 1
n
!1$
an = # & , n ' 1
"2%
cn =
3n!1 !1
, n "1
3n
n
!1$
gn = 9 # & , n ' 1
" 10 %
pn =
3n !1
, n "1
3n
n'1
!1$
rn = 2 # & , n ( 1
" 3%
tn =
4 n+1
, n !1
5n
the limit of the partial sums, as n grows without bound, is not a
finite value, The series diverges.
S = lim Sn = " . Because
n!"
$ 2n 1 '
$
2 n #1
1'
= lim && n # n )) = lim &1# n ) = 1# 0 = 1 = S .
n
n!" 2
n!" % 2
2 ( n!" % 2 (
S = lim Sn = lim
n!"
Because the limit of the partial
sums, as n grows without bound, is a finite value, the series S converges to 1.
Each new term in the sequence cn adds approximately 1/3 to each new partial sum in the
sequence Sn. Thus, the partial sums, Sn, continue to grow larger and larger without bound.
Because there is no upper bound for these partial sums, the series associated with the
sequence cn does not take on a finite value. The series S diverges .