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Steven Prascius
Mrs. Tallman
AP Calculus
31 March 2014
Sequences and Series
In calculus, sequences and series are two important topics that allow for the
approximation of functions and much more. Although sequences and series may seem to be the
same thing, they are quite different from one another. A sequence is an infinite ordered list of
numbers whereas a series is the summation of the numbers in a sequence. In other words, a
sequence is list of numbers and a series is the sum of that list of numbers. So if given a finite list
of the numbers one through five the sequence and series would be:
Sequence: 1, 2, 3, 4, 5
While the series would be
Series: 1+2+3+4+5
As seen in the example above, sequences and series both use the same terms but differ in that a
sequence merely list the terms in a list and a series takes the sum of the list of terms. Both
sequences and series have terms that follow the pattern t1, t2, t3….tn ,…where the t1 is the first term,
t2 is the second and so on with tn being the nth term in the list, with the sequence generally written
as an. Since series are sums of sequences, they are written as ∑𝑛𝑛=1 (an) where ∑ means
summation and n=1 and n are the first and last terms of the sequence an.
One thing sequences and series have in common is how they either converge or diverge.
When a sequence converges it means that as the terms in the sequence approach infinity the
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sequence will approach a single number, or limit. This is written as lim 𝑎𝑛 = 𝐿 where an is the
𝑛→∞
sequence and L is the limit, or number, that the sequence approaches as it goes toward infinity.
An example of a converging sequence is:
15
7
3
1
Converging sequence: 2, 116, 18, 14, 12………
As you can see the sequence is said to converge because the terms all approach 1 as they move
toward infinity. On the other hand, if a sequence does not converge it is said to diverge. When a
sequence diverges it means that it does not approach a single value as it goes to either positive or
negative infinity. This is written as lim 𝑎𝑛 = ∞. An example of a diverging sequence is:
𝑛→∞
Diverging sequence: 2, 4, 6, 8, 10, 12, 14……
As you can see the sequence is said to diverge because the terms do not approach a single value
as they move toward infinity, they just keep on going. A series converges if and only if its
sequence of partial sums converges. This means that if the terms in a sequence converge for a
certain range then the series of that sequence converge as well. If not then the series is said to
diverge. Usually a series diverges or converges depending on whether the sequence converges
or diverges, but there are situations when a sequence converges but the series diverges. One
such example is the harmonic series 1/n. When this series is written out as the sequence 1, 1/2,
1/3, 1/4... it is clear that it converges because the terms approach zero. The series, however,
diverges because the terms when added together approach infinity. This can be seen in figure 1.
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Figure 1. Harmonic series
Figure 1 shows how the series (in blue) diverges while the sequence (in red) converges.
Relating to sequences and series are Taylor and Maclaurin series. Taylor and Maclaurin
series allow any function f(x) to be written as an infinite sum of terms around a single point, x.
Taylor and Maclaurin series are calculated in the same way, with the derivative of the function
f(x) being taken several times and then having an x-value substituted into those derivative to get
a single f(x) value. This f(x) value is then substituted into Taylor series expansion form which
is:
Where f(a), f’(a), f’’(a), and f(3)(a) are the original function and its derivatives at the point a. This
point a is equal to a point x on the function. In the case of a Maclaurin series, this a value is
always equal to zero whereas in a Taylor series this a value can be any x-value. In short, Taylor
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and Maclaurin series are almost the same thing with the only difference being that a Maclaurin
series has to have a = 0. An example of a Taylor series can be seen when considering the
function sin(x) about x=π/3. In order to calculate the coefficients of the Taylor series, the
derivatives of sin(x) must be taken at x=π/3 which comes out to be:
f(x) = sin(x)
f(π/3) = √3/2
f’(x) = cos(x)
f’(π/3) = 1/2
f’’(x) = -sin(x)
f’’(π/3) = −√3/2
f(3)(x) = -cos(x)
f(3)(π/3) = -1/2
Once these derivatives are taken about x=π/3 they are plugged in as coefficients of the Taylor
series which comes out to be:
Sin(x) = √3/2 + 1/2(x-π/3) +
−√3/2
−1/2
2!
3!
(x- π/3)2 +
(x- π/3)3+……
Finding the Maclaurin series of the same function at x=0 would come out to be:
f(x) = sin(x)
f(0) = 0
f’(x) = cos(x)
f’0) = 1
f’’(x) = -sin(x)
f’’(0) = 0
f(3)(x) = -cos(x)
f(3)(0) = -1
Sin(x) = (x) +
−1
3!
(x) 3+……
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Another thing series can do is approximate the values of functions. A series can
approximate a value of a function by taking a partial sum at the x-value of that point. A partial
sum is simply the sum of a finite number of terms in a series. For example we could use the 5th
partial sum (5 terms) of the power series for ex to approximate e0.1. Since ex is one of the eight
basic power series, it is known that its expansion is:
1
n
ex = 1 + x + ½!*x2 + 1/3!*x3 + ¼!*x4 + ……..=∑∞
𝑛=0 𝑛! ∗ 𝑥
In order to approximate e0.1 we would simply plug 0.1 into the above expansion for x which will
give us an approximate answer for e0.1. When we do this it comes out to:
1
e0.1 = 1 + 0.1 + ½!*0.1 2 + 1/3!* 0.1 3 + ¼!*0.1 4 + ……..=∑4𝑛=0 𝑛! ∗ 0.1 n = 1.105170833
This approximation gives a value that is very close to the true value of e0.1 which is
1.105170918. The difference between the actual value and the approximate value is called the
error. In this situation the error is equal to 0.000000085. Since the error is so small it shows that
partial sums are an effective way of estimating the value of functions at certain points. This error
in the above problem can be lessened by increasing the number of terms, n, used in the partial
sum. The error calculated in the above example is known as the actual error and it is one of three
ways to calculate error and it again simply involves subtracting the approximate value from the
actual value.
Another way to calculate error is by the alternating series method. The alternating series
method states that the actual error will be no more than the absolute value of tn+1. The
alternating series method only applies to functions which pass the alternating series test, which
means that they converge. In order for a function to pass the alternating series test it must have
terms that alternate signs, have tn+1 < tn, and have the limit as n approaches infinity equal to zero.
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n
One such series that passes the alternating series method applies to is ∑∞
𝑛=2(−1) * 1/ln(n). This
series passes the alternating series test with each terms alternating signs, the limit as n
approaches infinity equaling zero, and having the tn+1 term less than tn.
The final way to calculate the error is the Lagrange method. The Lagrange method works
by finding the remaining terms of a series, or tail, to estimate the accuracy of the estimated
answer. This method states that the actual error must be less than or equal to the value of
𝑀
|𝑥
(𝑛+1)!
− 𝑎|n+1, with M being the maximum value of the functions derivative, n being the
number of terms, and x being the value that the estimate is being taken around. All three of these
methods accurately give the error between the estimated value of a function from a partial sum
and the actual value.
Sequences and series are two important topics in calculus. They allow for the expansion
of functions into individual terms and even allow values of functions to be approximated
accurately.
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5. Solve the following questions:
The function f is defined by the power series
n
f(x) = 1 + (x+1) + (x+1)2 +…+ (x+1)n +….= ∑∞
𝑛=0(𝑥 + 1)
for all real numbers x for which the series converges.
a) Find the interval of convergence of the power series for f. justify your answer.


(𝑥+1)𝑛+1
lim (|
𝑛→∞
(𝑥+1)𝑛
|) = |𝑥 + 1|, so -1 < x + 1< 1, therefore -2 < x < 0
n
When x = -2 you get ∑∞
𝑛=0(−1) which diverges by the nth term test because the
limit as n approaches infinity does not equal 0.

n
When x = 0 you get ∑∞
𝑛=0(1) which diverges by the nth term test as well
because the limit as n approaches infinity does not equal zero.

The interval of convergence is -2 < x < 0
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
F(x) = ∑∞
𝑛=0(𝑥 + 1) = 1−(𝑥+1) = −𝑥 in the interval -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

−0.5 1
g(-0.5) = = ∫−1
−𝑥
−0.5
𝑑𝑥 = -ln|𝑥| |
= 0.6931
−1
d) Let h be the function defined by h(x) = f(x2-1). Find the first three non-zero terms and
the general term of the Taylor series for h about x=0. Find the value of h(0.5).

h(x) = f(x2-1) = 1+ x2 + x4 +…….x2n +……

n
h(0.5) = f(-3/4) = ∑∞
𝑛=0(−3/4 + 1) = 4/3
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6. Which of the following series diverge? Be sure to address which test you have chosen and
WHY you chose it.
𝑛1.5 +1
a) ∑∞
𝑛=0( 5𝑛2 +7 )
𝑛1.5 +1
1
∞
2
∑∞
𝑛=1( 5𝑛2 +7 ) < ∑𝑛=0(𝑛2 ) , this series converges because it is bound by the series 1/n
which converges itself by the p-series test. I chose this test because it was easier than any
other test that could be done.
n
b) ∑∞
𝑛=2(−1) *
1
ln(𝑛)
This series converges by the alternating series test. The series alternates, has a limit that
1
1
approaches zero as n approaches infinity, and has ln(𝑛+1) < ln(𝑛). Because this series
fulfills all of these requirements it converges by the alternating series test. I chose this
test because it was obvious that the alternating series test would work.
4
n
n
c) ∑∞
𝑛=0(−1) *( 3)
4 𝑛/𝑛
4
n
n
∑∞
𝑛=0(−1) *( 3) = lim ( 3)
𝑛→∞
= 4/3 which is greater than 1
This series diverges by the root test. I chose the root test because the series had a root in
it and I thought that it would be easiest to do the root test.
7. Find the interval of convergence. Be sure to check the endpoints.


lim |(
𝑛→∞
2𝑥 𝑛+1
𝑛+2
2𝑥𝑛
)/(𝑛+1)| = |2𝑥| so -1/2 < x < 1/2
n
When x =-1/2 you get ∑∞
𝑛=0(−1) /n+1 which converges by the alternating series test.
The alternating series test applies because the series alternates signs, 1/n+2 < 1/n+1, and
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the limit as n approaches infinity is zero. I chose to the alternating series test because it
was apparent that the series fit that test the best.

n
When x = ½ you get ∑∞
𝑛=0(1) /n+1 which diverges by the limit comparison test when
compared to 1/n. This written out comes out to be lim (1n/n+1)/(1n/n) = lim (n/n+1) =
𝑛→∞
𝑛→∞
1. Since the limit is equal to a positive real number both series behave the same so since
n
1/n diverges by the p-series test so does ∑∞
𝑛=0(1) /n+1.

The interval of convergence is -1/2 <= x < 1/2
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Works Cited
Swift. "MAT 137, Calculus II." MAT 137, Professor Swift. N.p., n.d. Web. 30 Mar. 2014.
"Taylor Series." Wikipedia. Wikimedia Foundation, 30 Mar. 2014. Web. 30 Mar. 2014.