Gohlke 1
Ryan Gohlke
AP Calculus
Mrs. Tallman
13 April 2015
Sequences and Series
"Sequences and series are useful in the same way geometry is useful. There are things in
the world that can be represented by circles and squares, and things that can be represented as
sequences and series" (Leiviska). They are useful for approximations of functions, which is
necessary in virtually all computer applications involving evaluations of functions. Converging
and diverging sequences and series are important to understand, as well as error in the series
compared to the true value.
A sequence is an infinite ordered set of numbers. Example: 2, 3, 5, 7, 11, 13, 17, 19, 23,
... The sequence of primes. A series is the indicated sum of the terms of a sequence. Example: 2
+ 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + ... The series of primes. Sequences and series are
sometimes mixed up because they are closely related to each other. When talking about a
sequence there is no concern with the sum of the values of the terms because it is just looking at
the ordered set, whereas in a series there is interest in the sum of the ordered numbers. Geometric
and Arithmetic series are the two most common. They come in two forms, an explicit formula or
a recursive formula. Explicit there is no need for the previous term, but in recursive formulas,
there is a need for the previous term.
A sequence converges to L if its nth term approaches a finite limit L as n approaches
infinity. If the sequence does not converge, it is said to diverge.
Examples: 1, 1.5, 1.75, 1.875, 1.9375, ... converges to 2.
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2, 3, 5, 7, 11, 13, ... (the sequence of primes) diverges.
1, 0, 1, 0, 1, 0, 1, ... diverges by oscillation
A series converges if and only if its sequence of partial sums converges, otherwise the series
diverges. Example: 1 + 0.5 + 0.25 + 0.125 + 0.0625 + ... converges because its sequence of
partial sums is 1, 1.5, 1.75, 1.875, 1.9375, ..., which converges to 2. There are many tests to
determine if the series converges or diverges. It is possible for a sequence to converge and have
1
the same sequence diverge. For example, the Harmonic series ∑∞
𝑛=1 𝑛. The sequence converges
to zero, but the series when the terms are kept adding, then the series diverges.
In order to derive a Taylor and Maclaurin series, find the output of the function's
derivatives expanded about x = a. The coefficients should be C0 = f(a), C1 = f'(a), C2 =
=
𝑓′′(𝑎)
2!
,
C3
𝑓′′′(𝑎)
3!
, ... The powers are (x ─ a)n rather than xn because (x - a) equals zero when x = a. This
conforms to the formal rule of the Taylor and Maclaurin Series. If f is a function with
differentiable derivatives, then you can write f(x) as a Taylor series expansion about x = a as
follows: 𝑓(𝑥) = 𝑓(𝑎) + 𝑓 ′ (𝑎)(𝑥 − 𝑎) +
𝑓 ′′ (𝑎)
2!
(𝑥 − 𝑎)2 +
𝑓 ′′′ (𝑎)
3!
(𝑥 − 𝑎)3 +
𝑓 (𝑛) (𝑎)
𝑛!
(𝑥 −
𝑎)𝑛 + … If a = 0, the series is called a Maclaurin series, and a partial sum of a Taylor series is
called a Taylor polynomial. Basically the Taylor and Maclaurin series are the same except
Maclaurin pointed out the obvious and centered it about zero. Here is an example of a Taylor
1
Series: ln(𝑥) = (𝑥 − 1) − 2 (𝑥 − 1)2 +
1
1
3
Maclaurin Series: ex = 1 + 𝑥 + 2! (𝑥)2 +
(𝑥 − 1)3 −
1
1
(𝑥)4 + …
3!
(𝑥)3 +
1
4!
4
(𝑥 − 1)4 + … and an example of a
The use of a series to approximate values of functions is very helpful when the function
of ex, for example, is not working on a calculator and is on the test in which the question asks
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what is e0.1. So knowing the series to calculate values is important. There are three types of
errors: actually computing the error, alternating series, and the Lagrange Error.
An example of actually computing the error would be to find e0.1 and subtract it from a
partial sum of the ex series. (Refer to Problem 4).
The Alternating Series Estimation Theorem states that if ∑(−1)𝑛 𝑏𝑛 is a convergent
1
𝑛
alternating series, by AST, then |𝑅𝑛 | = |𝑠 − 𝑠𝑛 | ≤ 𝑏𝑛+1. For example, ∑∞
𝑛=1(−1) n this
1
1
1
1
1
1
converges by AST. Expand the series to 1 − 2 + 3 − 4 + 5 − 6. Add up the first three terms to 5/6
and it should be correct to at most the error first “neglected” term which would be 0.25 and that
is the term the AST Theorem stops at.
The Lagrange error is that a Taylor series is given and x = a is a number in the interval of
convergence , then there is a number c between a and x such that the remainder, Rn, after the
partial sum, Sn, is given the Lagrange form of 𝑅𝑛 =
𝑓 (𝑛+1) (𝑐)
(𝑥
(𝑛+1)!
− 𝑎)𝑛+1 and if M is the
maximum value of |𝑓 (𝑛+1) (𝑥)| on the interval between a and x, then the Lagrange error bound is
𝑀
|𝑅𝑛 | = (𝑛+1)! |𝑥 − 𝑎|𝑛+1 . For example, estimating e2 using the 11th partial sum. 𝑆10 =
1
𝑛
∑∞
𝑛=0 𝑛! 2 = 7.38899470... Using the Lagrange form of the remainder to estimate the accuracy
9
of partial sum plug into equation |𝑅10 | = (11)! |2 − 0|11 = 0.0004617... This means, by counting
the zeroes, the actual value should be off by two to four decimal places. In this case it will be
three decimal place because comparing e2 = 7.38905609... by calculator to 7.38899470.. from
above.
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Here are four AP problems and solutions that could possibly be on the AP Exam.
Problem 1
The function f is defined by the power series
∞
2
𝑛
𝑓(𝑥) = 1 + (𝑥 + 1) + (𝑥 + 1) + ⋯ + (𝑥 + 1) + ⋯ = ∑(𝑥 + 1)𝑛
𝑛=0
a) Find the interval of convergence of the power series for f. Justify your answer.
𝑛
Using the ratio term test, ∑∞
𝑛=0(𝑥 + 1) = lim |
𝑛→∞
(𝑥+1)𝑛+1
(𝑥+1)𝑛
| = lim (𝑥 + 1). Set (x + 1)
𝑛→∞
between the standard interval of -1 to 1. So (-1 < x + 1 < 1) and then solve for x to get the actual
interval of convergence of (-2 < x < 0). Check the endpoints, but both points of x = -2 and 0
𝑛
when ∑∞
𝑛=0(𝑥 + 1) diverge by the alternating series test (AST) and nth term test. This means
that the interval is open for both points.
b) The power series above is the Taylor series for f about x = -1. Find the sum of the series for f.
1
1
𝑛
To find the sum, we know the series is geometric so ∑∞
𝑛=0(𝑥 + 1) = 1− (𝑥 + 1) = − 𝑥 in
which this satisfies the interval of (-2 < x < 0).
𝑥
c) Let g be the function defined by g(x) = ∫−1 𝑓(𝑡)𝑑𝑡. Find the value of g(-0.5), if it exists, or
explain why g(-0.5) cannot be determined.
Notice g(-0.5) is in the interval of convergence. To find g(-0.5) plug into integral of g(−0.5
1
0.5) = ∫−1 − 𝑥 𝑑𝑡 which when integrated equals -ln |x| and evaluated at x = -1 to -0.5. This
equals ln 2 = 0.693147...
d) Let h be the function defined by h(x) = f(x2 - 1). Find the first three nonzero terms and the
general term of the Taylor series for h about x = 0. Find the value of h(0.5) (Answer on next
page).
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Simple, h(x) = f(x2 - 1) = 1 + x2 + x4 + ... + x2n + ... So plug in x to f and h (0.5) =
1
1−0.25
=
4
3
Problem 2
Which of the following series diverge? Be sure to address which test you have chosen
and WHY you chose it.
a) ∑∞
𝑛=0
𝑛3/2 + 1
5𝑛2 + 7
1
𝑛
b) ∑∞
𝑛=2(−1) ln(𝑛)
4
𝑛
𝑛
c) ∑∞
𝑛=0(−1) (3)
a) There is an n in both the denominator and numerator so nth term test could be used, but the
limit would go to 1 which is indeterminable. So, using L'hopitals rules we can say lim
𝑛3/2 + 1
𝑛→∞ 5𝑛2 + 7
∞
. Take the derivative, lim
∞
𝑛→∞
3/2𝑛1/2
10𝑛
=
∞
. Take derivative again, lim
∞
3/3𝑛−1/2
𝑛→∞
10
=
∞
10
=
= ∞. The
series then diverges.
b) Because there is a (-1)n we know that the series alternates. So, use an alternate series test and
there must be three rules that need to be true. (1) it alternates, (2) the limit as n approaches
1
infinity of one of ln of n is equal to zero (∞), and (3) one over the ln(n+ 1) is less than one over
the ln(n) which is true. The series converges by AST.
c) You can see that there is an indication of an alternating series, but when you run through the
4
three rules, the limit as n approaches infinity of (3)𝑛 = infinity so the series diverges. The nth
term test also says the ratio is greater than one so it then diverges.
Problem 3
Find the interval of convergence. Be sure to check the endpoints.
∑∞
𝑛=0
(2𝑥)𝑛
𝑛+1
= lim |
𝑛→∞
(2𝑥)𝑛+1
(𝑛+1)+1
(2𝑥)𝑛
𝑛+1
𝑛+1
|= lim | 𝑛+2 ×
𝑛→∞
1
(2𝑥)𝑛+1
(2𝑥)𝑛
| = 2𝑥 (−1 < |2𝑥| < 1). Solve for x so that
1
the interval of convergence is between − 2 < 𝑥 < 2. Now to check the endpoints, plug the
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endpoints into the series. When -0.5 is plugged into the equation. It has an alternating clue, so
use the alternating series test. (1) It alternates, (2) the limit as n approaches infinity of one over n
1
+ 1 is equal to zero (∞), and (3) one over n + 2 is less than one over n + 1which is true for all n.
The series converges at x = -0.5. Now for 0.5, 1n is always equal to one. Using a limit
𝑛
comparison test to 1/n, we know diverges because of harmonic series, and
(2𝑥)
∑∞
𝑛=0 𝑛+1
= lim
𝑛→∞
1
𝑛+2
1
𝑛+1
= 1. By limit comparison test, if the ratio equals one then they behave the same so it diverges at x
1
1
= 0.5. This gives the final interval of convergence to − 2 ≤ 𝑥 < 2.
Problem 4
Write the 5th partial sum of the power series for ex. Use this partial sum to approximate
e0.1. Discuss the error in your estimate.
𝑆5 = 1 + 𝑥 +
𝑆5 = 1 + 0.1 +
𝑥2
𝑥3 𝑥4
+
+
+ …
2!
3!
4!
0.12
0.13 0.14
+
+
+ …
2!
3!
4!
S5 = 1.10517083334...
e0.1= 1.1051709180756...
Subtract the real value minus the partial sum to get 0.0000000847422 which is the error. It is
also known as the value of the tail of this series after the fifth term. The Lagrange error will give
similar results which 0.0000000847422 is going to be less than the Lagrange error.
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Works Cited
Leiviska, Tara. "Why Study Sequences and Series?" Dr. Math. The Math Forum, 26 Feb. 1999.
Web. 31 Mar. 2015. <http://mathforum.org/library/drmath/view/56618.html>.