Sequences: A look into interesting sequences, series and convergence
Introduction
Sequences have had an impact on mathematics. They have been used in many instances: it has been
used in teaching younger generations multiplication. There have also been many terms which have
been associated with sequences, such as nth term and partial sums. There has also been the creation
of many interesting sequences and series which have puzzled many mathematicians, such as the
Fibonacci sequences and Harmonic Series. Also, series with the same concept have different sums
and many test and theories have been published. In this essay, I will be looking into convergence and
divergence of interesting sequences and series.
To start off, I will be defining some important terms. First, the nth term of a sequence is an
expression which can be used to find a certain term in the sequence without having to go through all
previous terms. A partial sum is a sum of all terms in a sequence until a certain point. Also, the
difference between a sequence and series is that a sequence is a set of number with can be defined
by an nth term. However, a series is a sequence of partial sums of a sequence.
Convergence Test used in this essay
Like explained beforehand, a series is a sequence of partial sums of another sequence. There have
been many tests developed to help figure out whether a series is convergent or divergent. However,
there is not a single test which can guarantee an answer. For now, we will be looking at The
Divergence Test and The Ratio Test.
The Divergence Test goes as followed1
If the lim 𝑎𝑛 ≠ 0, or doesn’t exist,∑∞
𝑛=1 𝑎𝑛 diverges.
𝑛→∞
The Ratio Test goes as followed2
lim |
𝑛→∞
𝑎𝑛+1
|
𝑎𝑛
< 1 ⇒ ∑∞
𝑛=1 𝑎𝑛 converges
𝑎𝑛+1
|
𝑛→∞ 𝑎𝑛
lim |
> 1 ⇒ ∑∞
𝑛=1 𝑎𝑛 diverges
1. Fibonacci Sequence
To start off, I will be looking at the nth of a very important sequence, known as the Fibonacci
sequence. The general rule for this sequence is to add the previous 2 terms together. We define the
nth term of the Fibonacci sequence as
𝑢𝑛 = 𝑢(𝑛−1) + 𝑢(𝑛−2)
1
Oregonstate.edu. (2020). Divergence Test. [online] Available at:
https://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/SeriesTe
sts/divergence.html [Accessed 8 Jan. 2020].
2
Oregonstate.edu. (2019). Ratio Test. [online] Available at:
https://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/SeriesTe
sts/ratio.html [Accessed 8 Jan. 2020].
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Sequences: A look into interesting sequences, series and convergence
As we know that this sequence is not linear as there is no equal gap, we can safely assume that the
sequence is geometric and we can derive at the nth term of the sequence by replacing 𝑢𝑛 for 𝑘𝑥 𝑛 , so
the previous equation can be put on the form of
𝑘𝑥 𝑛 = 𝑘𝑥 𝑛−1 + 𝑘𝑥 𝑛−2
The next step is to simplify this equation down. We can do this by dividing the equation by 𝑘𝑥 𝑛−2, so
the equation would be
𝑥2 = 𝑥 + 1
Now we have a quadratic equation which we can solve by using the quadratic formula. There would
1+√5 1−√5
, 2 ).
2
be 2 possible solutions for 𝑥, (
So, if we put both into equations and we would have
𝑛
𝑢𝑛 = 𝑘1 (
1+√5
2
𝑛
) and 𝑢𝑛 = 𝑘2 (
1−√5
2
)
where k is some constant. As we don’t know which equation could be the answer, we are going to
put both equations into one new equation and changing the constant k into A and B,
𝑛
𝑛
1 + √5
𝑢𝑛 = 𝐴
1 − √5
+𝐵
2
(
)
2
(
)
Now we are going to solve for A and B. We also know that u0 = 0, u1 = 1. So using this we can solve
for A and B. Using the fact that u0 = 0, we get to the fact that 0 = 𝐴 + 𝐵 ∴ 𝐵 = −𝐴. This means we
can change the equation to
𝑛
𝑢𝑛 = 𝐴 (
1+√5
2
𝑛
) − A(
1−√5
2
) .
Now using the fact that u1 = 1, we can solve for A.
1 = 𝐴√5
→𝐴 =
1
√5
∴𝐵=−
1
√5
So now we have the nth term for the Fibonacci sequence
𝑛
𝑛
1 + √5
1 − √5
−
2
𝑢𝑛 = (
)
(
√5
2
)
This equation is known as Binet’s formula3. Now we can figure out whether the Fibonacci sequence
converges or diverges. Using the fact that the denominator will never change as there is no
unknown, the convergence and divergence of the sequence is dependent on whether the top half of
the sequence will ever reach infinity. Because the sequence is always increasing at a very high rate, it
3
Natividad, L. R. (2011). Deriving a formula in solving Fibonacci-like sequence. International Journal
of Mathematics and Scientific Computing, 1(1), 19-21.
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Sequences: A look into interesting sequences, series and convergence
has the possibility of reaching infinity and being unbounded. Therefore, the Fibonacci sequence is a
divergent sequence.
Now, we can look at whether the Fibonacci Series converges or diverges. First, we previously know
that lim 𝑎𝑛 = ∞. With this, we can use the Divergence test. This test shows us that it will diverge,
𝑛→∞
as the lim 𝑎𝑛 ≠ 0
𝑛→∞
2. Harmonic Sequence
Another interesting sequence is the Harmonic sequence. The Harmonic sequence is a sequence
which has the nth term
𝑎𝑛 =
1
𝑛
The sequence would go as followed
1 1 1 1
1, , , . , ⋯
2 3 4 5
To figure out whether the sequence is convergent or divergent, we will need to look at
1
lim
𝑛→∞ 𝑛
First of all, the numerator is always at 1, this shows that the numerator is bounded and
lim 1 = 1
𝑛→∞
Next, the denominator is n. As n is a variable in this expression, it has no limit/unbounded and can
go to infinity. However, as the denominator gets bigger, the number itself actually gets smaller. This
number also gets closer to 0 as the numerator does not change. To graph this, we can put the
1
equation as 𝑦 = 𝑥. This would look like the following.
This graph has 2 asymptotes, 𝑥 = 0 and 𝑦 = 0. This is where there is no solution of numbers which
can make these values true for this equation. This graph is known as a reciprocal graph. Also from
the graph, we can also tell that the value for y is getting closer and closer to 0 when we go either
1
way. Therefore, we can say that the Harmonic sequence converges at 0 because lim 𝑛 = 0.
𝑛→∞
Next, we will be looking at whether the Harmonic Series converges or diverges. This is the series for
the Harmonic Sequence. There are actually many proofs for this series, so first we are going to come
to a conclusion using the Divergence and Ratio tests, then we are going to look at the proof by
Nicole Oresme, which dates back to 1350.
First of all, we found out that the lim 𝑎𝑛 = 0. Therefore, according to the Divergence Test, it is not
𝑛→∞
a divergent series. However, this doesn’t confirm that the Harmonic Series is a convergent series.
Next, we are going to use the Ratio test to figure out whether the Harmonic Series is a convergent or
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Sequences: A look into interesting sequences, series and convergence
divergent sequence. So the equation we have to solve is
1
𝑛
+
1|
lim |
1
𝑛→∞
𝑛
this simplifies down to
𝑛
lim |
|
𝑛→∞ 𝑛 + 1
To find a value for this expression, we can use l’Hôpital’s Rule4 to simplify this down. This says that
when 2 functions have no bounds and can approach infinity, you can simplify the fraction down by
finding the derivative of both functions. So, we have to find the derivative of both the numerator
and denominator. The derivative of the numerator and demoninator are 1. Therefore, the simplified
form of the previous equation is
lim |1| = 1
𝑛→∞
However, this follows neither of the 2 requirements for the Ratio Test, so we have not found
whether the Harmonic Series is convergent or divergent.
However, there is a proof for the divergence of the Harmonic Series. This proof was created by
Oresme. It goes as followed:
∞
Consider the subsequence {𝐻2𝑘 }𝑘=0
𝐻1 = 1 = 1
𝐻2 = 1 +
1
1
=1+
2
2
1
1 1
1 1
𝐻4 = 1 + + ( + ) > 1 + + = 2
2
3 4
2 2
1
1 1
1 1 1 1
1 1 1
1
𝐻8 = 1 + + ( + ) + ( + + + ) > 1 + + + = 2 +
2
3 4
5 6 7 8
2 2 2
2
In general,
1
𝐻2 𝑘 ≥ 1 + 𝑘 ( )
2
Since this subsequence is unbounded, the Harmonic series must also diverge. Using this type of
𝑀−1
).
𝑀
argument, one can show that for any positive integer M, 𝐻𝑀𝑘 ≥ 1 + 𝑘 (
This argument has no flaws as each step makes sense and the conclusion makes no assumptions and
therefore the conclusion must be true.
3. p-Harmonic Sequence
Another set of interesting sequences are the p-Harmonic Sequences. These sequences take the same
form as the Harmonic sequence but the denominator has a different power to it. We will look at the
1
p-Harmonic Sequence when p is 2. If p=2, the nth term for this sequence is 𝑏𝑛 = 𝑛2 . Using the same
logic as we did for the Harmonic sequence, the numerator is a fixed value and is bounded to 1 and
the denominator is an unknown so it can go to infinity. Also as the denominator gets larger, the
4
Oregonstate.edu. (2020b). l’Hopital’s Rule. [online] Available at:
https://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/s
tatement.html [Accessed 8 Jan. 2020]
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Sequences: A look into interesting sequences, series and convergence
value of the fraction becomes smaller. If we graph this sequence with the equation 𝑦 =
1
,
𝑥
1
,
𝑥2
we will
have the same two asymptotes as 𝑦 = x=0 and y=0. Also, the graph showing that the numbers
keep getting closer and closer to 0 as it gets further away from the origin.
Next, we will look at whether the p-Harmonic Series converges or diverges, when p=2. Instead of
using the Ratio and Divergence Test, we are going to adapt Oresme’s proof for the divergence of the
1
Harmonic Series. This time instead of the numbers being 2, it is going to change and get smaller. We
are going to make a sequence called 𝑘𝑛 , is sequence will go as followed,
1 1 1 1 1 1 1 1 1 1 1
1, 2 , 2 , 2 , 2 , 2 , 2 , 2 , 2 , 2 , 2 , 2
2 2 4 4 4 4 8 8 8 8 8
If we compare this sequence with the p-Harmonic sequence, we notice that all the terms in the pHarmonic sequence are less than or equal to their corresponding term in this sequence. Also, this
sequence’s series will converge at 2 because each block of terms when added together always shrink
1
in size and it will create a sequence which has the nth term of 2𝑛 . This will never exceed 2. Therefore,
this series is convergent and converges to a value below 2.
4. Geometric Sequence
We will look at geometric sequences. Geometric sequences have an nth term which come in the
form:
𝑐𝑛 = 𝑎𝑟 𝑛
Sequences like have very large gaps, which widen as the variable which changes is the degree of
polynomial of r. This means the number is usually very large. When finding whether a geometric
sequence converges, we need to look at several cases: when 0 > |𝑟| > 1, | 𝑟| = 1, | 𝑟| < 1.
First, we will look at when |𝑟| = 1. If we express this an formula, we will get
𝑐𝑛 = 𝑎 ∗ 1𝑛
When 1 is to the power of anything, we get the same result. This means it will converge to a. As 1
multiplied by a is a.
Then, if |𝑟| > 1, r will increase at a rapid rate. When a number above one has its power raised, it has
a large increase. This means as r increases, it will have no limit as it will get larger and larger at a
large rate. Therefore, when |𝑟| > 1, the geometric sequence will diverge.
Finally, if 0 < |𝑟| < 1, we know that the value for 𝑟 𝑛 will decrease as n increase. This means the
value of the term will decrease. This also means it will not pass a certain number: this number is 0.
When a number below 1 has its power raised, it comes closer and closer to 0. Therefore, we know
that when 0 < |𝑟| < 1, it will converge to 0.
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Sequences: A look into interesting sequences, series and convergence
Now, we will look at whether geometric series converge or diverge or whether there are necessary
requirements for it to converge. We will use the tests listed at the start to figure this out. We will
also refer to the same three cases above |𝑟| > 1, | 𝑟| = 1, 0 < |𝑟| < 1.
First, we know that when |𝑟| > 1 the sequence diverges, Accoding to the Divergence Test, this series
can not converge and therefore must only be able to diverge.
Now, we are going to use the Ratio Test to figure out what are the requirements for convergence of
geometric series. For the term substitution, we will set 𝑎𝑛 as 𝑎𝑟 𝑛 and 𝑎𝑛+1 as 𝑎𝑟 𝑛+1 . When we
substitute these in, we get
𝑎𝑟 𝑛+1
|
𝑎𝑟 𝑛
lim |
𝑛→∞
If we simplify the fraction we end up with:
lim |𝑟|
𝑛→∞
This means for the series to converge r must be less than 1, according to the Divergence Test.
However, when 𝑟 = 1, the term will all be a and when you have a repeating sequence the sum will
always increase as the gap between turns are the same.
Therefore, geometric series only converge when 0 < |𝑟| < 1.
Summary
For summary, we have derived at a formula for the nth term of the Fibonacci sequence, which is
known as the Binet’s Formula:
𝑛
𝑛
1 + √5
1 − √5
−
2
𝑢𝑛 = (
)
(
√5
2
)
We also took a look at convergence and divergence in famous sequences. This was done why using
the nth term of a sequence and using basic arithemetic rules and a short look into polynomials.
Next, we looked at the respective sequences’ series and used tests to figure out whether they were
convergent or divergent. We had a look at Oresme’s proof for the divergence of the Harmonic
Series.
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Sequences: A look into interesting sequences, series and convergence
Bibliography
Kifowit, S., and T. Stamps. "The harmonic series diverges again and again." The AMATYC
Review 27.2 (2006)
Mathsisfun.com. (2017). Derivative Rules. [online] Available at:
https://www.mathsisfun.com/calculus/derivatives-rules.html [Accessed 8 Jan. 2020].
Natividad, L. R. (2011). Deriving a formula in solving Fibonacci-like sequence. International
Journal of Mathematics and Scientific Computing, 1(1), 19-21.
Oregonstate.edu. (2019). Ratio Test. [online] Available at:
https://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/Sand
S/SeriesTests/ratio.html [Accessed 8 Jan. 2020].
Oregonstate.edu. (2020a). Divergence Test. [online] Available at:
https://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/Sand
S/SeriesTests/divergence.html [Accessed 8 Jan. 2020].
Oregonstate.edu. (2020b). l’Hopital’s Rule. [online] Available at:
https://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/Sand
S/lHopital/statement.html [Accessed 8 Jan. 2020].
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