Lecture 2 – Appendix B: Some sample problems from Boas, Chapter 1
Here are some solutions to the sample problems assigned for Chapter 1.1, 1.6 and 1.9.
§1.1: 5
Solution: We want to use the general expression for the form of a geometric series

S   ar n 
n 0
a
1 r
To find the fraction corresponding to S  0.58333 . To use this expression we need
to look for a repeating decimal in order the geometric series. To this end we rewrite
the starting number as
S  0.583333  0.25  0.3  0.03  0.003  0.0003  .
Now we can rewrite the repeating decimal as a sum and use the general form to find
the desired fraction,
1
1
1
1
1


 0.3 1   2  3  4  
4
 10 10 10 10

1
0.3
1 0.3 1 1 7
 
 
   .
4 1  1 4 0.9 4 3 12
10
S
§1.1: 12
Solution: Water purification as defined here is another process described by a
geometric series. If one-nth of the current impurity is removed at each stage, the total
fraction of the impurity removed after N stages of purification is given by
1
1 1
1 1 1  nN
Impurity  n, N    2    N 
n n
n
n 1 1
n
1 nN 1
 N
.
n n 1
Physics 227 Lecture 2 Appendix B
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So now we consider the limit N   for the two cases mentioned, n = 2 and n = 3,
too find what fraction of the impurity we can remove in the limit of an infinite
number of stages,
1 2N  1
2N
lim N  Impurity  2, N   lim N  N
 lim N  N  1,
2 2 1
2
1 3N  1
3N 1 1
lim N  Impurity  3, N   lim N  N
 lim N  N  .
3 3 1
3 2 2
Thus in the first case all of the impurity can be removed, but in the second case, with
only a third removed at each stage, the best we can hope for is to remove one half of
the impurity.
§1.6: 3
Solution: We want to practice using the comparison test for convergence/divergence.
Let’s proceed by comparing to the convergent geometric series with x  1 2


1
1 1 1 1

1

     cn  2 n  .
n
2 4 8 16
n 0 2
C   cn  
n 0
Now we compare this to the series of interest,

1
1 1 1
 1       an  n 2 .
2
4 9 16
n 1 n
S 
As suggested in the problem we proceed by performing some manipulations, i.e.,
resuming, on the 2 series. We have
1
1
1   1
1 
1 1  1
S 1      

    
  ,
225 
 4 9   16 25 36 49   64
i.e., we look at the first term, then add together the next 2 terms, then add the next 4
terms, then add the next 8 terms, etc. Now compare this to a similar redefinition of
the known convergent series
Physics 227 Lecture 2 Appendix B
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1 1 1
  
2 4 8
2 4
8
1  

4 16 64
1 
1 1  1 1 1 1  
 1             8 terms of
  .
64 
 4 4   16 16 16 16  
C 1
Comparing these two forms for the two series shows that each parenthesis in series S
is smaller than the corresponding parenthesis in series C. Thus, since C converges,
the comparison test tells us that S must converge also.
§1.6: 16
Solution: This exercise is intended to encourage us to think about the (possibly
incorrect) role of the lower limit in the integral test. In considering the sum of terms
an  1 n2 we might be tempted to include a lower limit of zero and evaluate the
integral


dn
1


0 n2 n 0  0      .
This result is misleading since the infinity arises entirely from the lower limit. On the
other hand we know that the convergence/divergence of the infinite series cannot
depend on the first N terms of the series, where N is finite. By the same token the
convergence/divergence of the series cannot depend, i.e., be determined by, the
behavior of the integral at the lower limit. Clearly choosing any lower limit other
than the special value zero will lead to a finite integral and the correct conclusion that
the series converges. The lesson is that you should just ignore the behavior of the
integral at the lower limit. It tells us nothing useful!
§1.6: 20
Solution: Here we want to practice using the ratio test. With the specified series we
have
Physics 227 Lecture 2 Appendix B
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
n!
n!
, an 
,
2
n
!
2
n
!




n 0
S 
n 
an 1  n  1 !  2n  !
n 1
1



an
n !  2n  2  !  2n  2  2n  1 2  2n  1
  lim n  n  lim n
1
 0.
2  2n  1
Thus the ratio test tells us that the series converges (absolutely).

nn
*§1.9: 9 Here we consider the series  . First look at the preliminary test,
n 1 n !
nn
nn
en
limn an  limn  limn n  n
 limn
 .
n!
n e 2 n
2 n
Thus the series must diverge. To see this form of the asymptotic behavior of the
factorial function look ahead to Eq. (11.1) in Chapter 11 (Stirling’s formula).
2   1
*§1.9: 16 The series of interest is defined by  2
. We first note that the
n 0 n  7
 2  1 , vanish as n   and the preliminary test is passed.
individual terms, an 
n2
Next consider the ratio test which says
n

  limn
2   1
n 1
2   1
n
n2  7
 n  1
2
7
.
Due to the behavior of the first factor it is somewhat difficult to evaluate the large n
limit. However, since this factor is either 3 or 1, when can proceed to consider the
appropriate compassion series given by


2   1
1
3


.



2
2
2
n

7
n

7
n

7
n 0
n 0
n 0
n

Now apply the ratio test to the larger series to find
Physics 227 Lecture 2 Appendix B
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3 n2  7
1  7 n2
 2
  limn
 limn
 limn 1   .
2
2
3  n  1  7
1 2 n  8 n
 n
Now we can easily apply the rule in Eq. (2.29) to see that this series, and thus the
series of interest, converges (p = 2 > 1) .
Physics 227 Lecture 2 Appendix B
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