MIDTERM 2 STUDY GUIDE.
MARTIN OLSSON
Disclaimer: This document is intended as an aide in studying for the exam. Not everything
on this sheet will be on the exam, and conversely there may be questions on the exam not
covered on the sheet. The complete list of things covered is specified in the syllabus.
This document will be amended to include the last few lectures before the exam next week.
So far this is my study suggestions for the sequences and series component of the midterm.
§1 Getting started:
To get oriented I recommend doing the following:
1. Go over each section covered in the book, as well as your lecture notes. Be aware that
sometimes things are covered in the book that was not discussed in lecture, and sometimes
the lecture discussion had more information than what was in the book.
2. Review basic definitions. With all the theory, it is easy to forget the basics. For example,
what does it mean for a sequence or series to converge?
3. Make sure to review sequences! So much attention has been focused on the series
chapters but sequences are certainly also part of the exam.
4. For the various tests for convergence of a series, make flashcards for each of the tests
and make sure you understand the assumptions that go into each test. A typical source of
mistakes in this regard is that students try to apply tests without verifying the assumptions
and then get incorrect answers.
5. Make sure you understand the picture I drew in lecture concerning intervals of convergence for power series.
6. Memorize the table on page 743 of the book for the various standard power series and
their radii of convergence.
7. Spend some time thinking about what other power series you can generate from the
series in 6 by integration, differentiation etc.
8. Study the section on the remainder for Taylor series and how one uses it to prove that
a function is equal to its Taylor series.
9. Go over homework problems making sure that you can do the basic problems for each
section.
§2 Take practice exams.
1
2
MARTIN OLSSON
The heading for this section says it all. Take some practice exams from the exam archive.
I will choose a specific one for the review session, but once you have the basic understand
take an exam to see how it goes. Make sure to mimic the exam environment as closely as
possible (i.e. time yourself for 50 minutes, and use no outside aids).
After taking an exam try to identify weak areas and go back to section 1.
§3 Some questions for study.
Below are some questions some of which I have given in previous 1b classes. Not all of the
following problems are suitable as exam questions but I think they are all good to study. You
should also study the homework problems and other problems from the book. The problems
here are intended as supplements.
1. (a) Give an example of a sequence of positive numbers that is bounded above but does
not converge.
(b) Give an example of a sequence that is monotonically increasing but does not converge.
2. Repeat the previous exercise for each of the standard tests for convergence (integral test,
ratio test etc.). Remove one of the assumptions of the test and give a counterexample.
3. (I asked this in a previous 1b class) Determine if the following statement is true or false:
P∞
P
n
If ∞
n=1 (an ) converges.
n=1 an converges, then
If true explain why, if false give a counterexample.
Solution: The statement is true. The easiest way to see this is to use the root test. Let bn
denote (an )n . We compute
p
lim n |bn | = lim |an |
n→∞
n→∞
P∞
which
P∞ is zero because n=1 an converges. Since 0 < 1 we conclude by the root test that
n=1 bn converges absolutely, and in particular converges.
P
Alternate solution: Since ∞
n=1 an converges, we know that limn→∞ an = 0, which means
n
n
in particular that there is an
N
P∞ such thatn |an | < 1/2 for all n > N . Therefore |an | < (1/2)
for all n > N , and sinceP n=N +1 (1/2) converges (it is a geometric series), it follows by
∞
n
the comparison test that P
absolute convergence implies
n=N +1 |an | converges as well.
PSince
∞
∞
n
convergence, we have that n=N +1 an converges. Thus n=1 an converges.
4. (I asked this in a previous 1b class)
(a) Does
∞
X
n=1
10n
n · 3n+1
converge or diverge? Prove your answer.
MIDTERM 2 STUDY GUIDE.
3
Solution: There were several ways to do this problem. One way is to do a limit
comparison test (or comparison test) with bn = 1/n (note that the terms in the series
are non-negative). We get
an
1 10
= lim 10n /3n+1 = lim ( )n .
n→∞
n→∞ 3 3
n→∞ bn
P
This limit goes to ∞ since 10/3 is bigger than 1. Since ∞
n=1 (1/n) diverges (Harmonic
series) by the limit comparison test stated in class (more general than the one in the
book) the sum in question also diverges.
(b) Does the series
∞
X
3
lim
n=1
n2 + 5n + 4
converge or diverge? Prove your answer.
Solution: Here the best seems to be to use limit comparison test (note again the
terms are nonnegative) with bn = 1/n2 . If an denotes 3/(n2 +5n+4) the limit becomes
an
3
3n2
= lim
= 3.
= lim 2
n→∞ bn
n→∞ 1 + 5/n + 4/n2
n→∞ n + 5n + 4
P∞
P
Since 3 > 0 and ∞
n=1 an
n=1 bn converges (p-series with p = 2) it follows that
converges as well.
(c) Bonus: (Try this problem only if you have completed all the other problems). Compute the sum of the series in (b) (hint: consider the partial fraction decomposition).
Solution: Using partial fractions one sees that
lim
n2
3
1
1
=
−
.
+ 5n + 4
n+1 n+4
Unfortunately neither of these series converges by themselves so we cannot
P treat them
3
separately. However, note that if we compute the n-th partial sum sn of ∞
n=1 n2 +5n+4
we get after reindexing and for n at least 3
sn =
n+1
X
1
i=2
i
−
n+4
X
1
i=5
i
=
1 1 1
1
1
1
+ + −
−
−
2 3 4 n+2 n+3 n+4
since all the middle terms cancel. Therefore the limit of the sn ’s is equal to
1 1 1
1
1
1
1 1 1
13
lim sn = lim ( + + −
−
−
)= + + = .
n→∞
n→∞ 2
3 4 n+2 n+3 n+4
2 3 4
12
5. (I asked this in a previous 1b class). (a) Find the radius of convergence of the power
series
∞
X
(n!)2 n
x .
(2n)!
n=1
(b) Determine whether the power series converges or diverges at the end points of the
interval of convergence.
4
MARTIN OLSSON
Solution: (a) Use the ratio test:
lim |
n→∞
((n + 1)!)2 n+1 (2n)! 1
|x|
n→∞ (2n + 2)!
(n!)2 |x|n
(n + 1)2
= lim
|x|
n→∞ (2n + 2)(2n + 1)
(n2 + 2n + 1)|x|
= lim
.
n→∞ 4n2 + 6n + 2
an+1
| =
an
lim
Since n2 + 2n + 1 and 4n2 + 6n + 2 are both polynomials of degree 2, we have
(n2 + 2n + 1)|x|
1
= |x|.
2
n→∞ 4n + 6n + 2
4
lim
Therefore the series converges if and only if 14 |x| < 1. Equivalently |x| < 4. Therefore the
radius of convergence is 4.
(b) At the endpoints we get the two series
∞
X
(n!)2 4n
n=1
(2n)!
,
∞
X
(−1)n (n!)2 4n
(2n)!
n=1
.
I claim that both series diverge. By the test for divergence it suffices to show that the limit
(n!)2 4n
n→∞ (2n)!
lim
is not equal to zero. For this write
(n!)2 4n
(n)2 (n − 1)2 (n − 2)2 · · · (2)2 (12 )4n
(2n)2 (2n − 2)2 (2n − 4)2 · · · (2 · 2)2 (2)2
=
=
.
(2n)!
(2n)(2n − 1)(2n − 2) · · · (4)(3)(2)(1)
(2n)(2n − 1)(2n − 2) · · · (4)(3)(2)(1)
We can write this as
(2n − 2)(2n − 2)
(4)(4) 2 · 2
(2n)(2n)
·
···
·
.
(2n)(2n − 1) (2n − 2)(2n − 3)
4·3 2·1
Each of the individual fractions is > 1, so we find that
follows that
(n!)2 4n
lim
6= 0
n→∞ (2n)!
so the series diverges at both endpoints.
(n!)2 4n
(2n)!
is bigger than 1 for all n. It
6. (I asked this in a previous 1b class.) (a) Find the Maclaurin series of f (x) = x2 e−x .
(b) State Taylor’s inequality (explaining carefully in complete sentences the statement of
the theorem and the meaning of the various terms in the formula).
Solution: (a) First recall from class that
ex =
∞
X
xn
n=0
n!
.
MIDTERM 2 STUDY GUIDE.
Therefore
−x
e
=
∞
X
(−x)n
n=0
n!
=
∞
X
(−1)n xn
n!
n=0
5
.
Multiplying this series by x2 we get
x2 e−x =
∞
X
(−1)n xn+2
n=0
n!
.
(b) Let f be a differentiable function, and let Tn (x) be the n-th Taylor polynomial centered
at c
n
X
f (i) (c)(x − c)i
Tn (x) =
.
i!
i=0
The error is then defined to be
Rn (x) = f (x) − Tn (x).
Let d be a number and M a number such that |f (n+1) (x)| ≤ M for |x − c| ≤ d. Then the
remainder satisfies
M
|Rn (x)| ≤
|x − c|n+1
(n + 1)!
for |x − c| ≤ d.
7. Let f be a function with domain (0, ∞), and suppose there exists a power series
∞
X
an x n
n=0
such that for all t in the interval (0, ∞) we have
∞
X
f (t) =
an tn .
n=0
Show that there exists a continuous function F defined on all real numbers such that F (t) =
f (t) for t ∈ (0, ∞).
Solution: Since the power series is centered at 0 and converges for all t > 0, the radius of
convergence must in fact be ∞. Therefore the power series defines a continuous function on
all reals which agrees with f on (0, ∞).
8. Does the series
∞
X
√
n+1−
√
n
n=0
converge? How about
∞
X
n=0
Solution:
√
n+1−
n
√
n
6
MARTIN OLSSON
Write
√
√ √
√
√
√
( n + 1 − n)( n + 1 + n)
n+1−n
1
√
n+1− n=
=√
√
√ =√
√ .
n+1+ n
n+1+ n
n+1+ n
√
Therefore by limit comparison test with 1/ n we get that the first series diverges.
Similarly we get that the second series converges by limit comparison test with 1/n3/2 .
9. Let {an } be a sequence of positive integers, infinitely many of which are nonzero. Show
that the radius of convergence of the power series
∞
X
an x n
n=0
is at most 1.
Solution: If |x| ≥ 1 then for every nonzero an we have an |x|n ≥ 1 since the an are integers.
Since infinitely many an ’s are nonzero we conclude that limn→∞ an |x|n 6= 0 if |x| ≥ 1 so the
series diverges for |x| ≥ 1.
10. (a) By multiplying series, show that the Maclaurin series for the function
f (x) = sin(x) cos(x)
is equal to
∞
X
an x2n+1 ,
n=0
where
2n+1
X 2n + 1
(−1)n
an =
.
i
2((2n + 1)!) i=0
(b) Using the half angle formula show that
∞
1X
x2n+1
(−1)n 22n+1
.
sin(x) cos(x) =
2 n=0
(2n + 1)!
(c) Combining (a) and (b) deduce the identity
2n+1
X 2n + 1
= 22n+1 .
i
i=0
Can you prove this formula directly from the definition of the binomial coefficient?
Solution: (a) Recall that
sin(x) = x −
x3 x5
x2 x4
+
− · · · , cos(x) = 1 −
+
− ··· .
3!
5!
2!
4!
MIDTERM 2 STUDY GUIDE.
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Multiplying these two series we see that we get only odd exponents, and that the coefficient
of x2n+1 is
1
1
1
1
(−1)n ( (2n)! +
+
+ ··· +
).
1!
3!(2n − 2)! 5!(2n − 4)!
(2n + 1)!0!
Now recall that
2n + 1
(2n + 1)!
2n + 1
=
=
.
i
i!(2n + 1 − i)!
2n + 1 − i
We conclude that the coefficient can be written as
2n+1
X 2n + 1 (−1)n
2((2n + 1)!)
i=0
i
.
(b) The half angle formula gives that
1
sin(2x).
2
Therefore plugging 2x into the Taylor series for sin(x) we get that
∞
x2n+1
1X
(−1)n 22n+1
.
sin(x) cos(x) =
2 n=0
(2n + 1)!
sin(x) cos(x) =