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10.1 Defining Convergent and Divergent Infinite Series

2 min readjune 18, 2024

Avanish Gupta

Avanish Gupta

Avanish Gupta

Avanish Gupta

10.1 Defining Convergent and Divergent Infinite Series

Welcome to unit 10 of AP Calculus BC! This is the last one. 🥳

This first, longer part of the guide concerns the behavior of numerical series and whether they have an actual sum as the number of terms approaches infinity. There are many tests and methods that we can use to answer this question, each working on specific cases. However, before we can answer this, we need to first talk about what sequences and series are.

Before we start talking about series, we need to talk about sequences and some terminology about sequences.

🤔 What is a Sequence?

A sequence is just a list of terms related by a common pattern to each other.

Here is how we represent sequences:

{an}1\{a_n\}^\infty_1

📝 Terms in a Sequence

List a1,a2,a3,ana_1, a_2, a_3, a_n, and an+1a_{n + 1} for the following sequences. We can go through this together!

✏️ Finding Terms in A Sequence: Example 1

{1n}1\{\dfrac1n\}^\infty_1

This is nothing more than just plugging in n = 1, 2, 3, n, and n+1 into ana_n.

{1,12,13,...,1n,1n+1,...}\{1, \dfrac12, \dfrac13,...,\frac{1}{n}, \frac{1}{n+1}, ...\}

We use the ... between the 3rd and the nthn^{th} term and after the (n+1)th(n+1)^{th} term because there are an indefinite number of terms in these intervals.

This is a special sequence called the harmonic sequence. We will learn more about the following study guide later in this unit: 10.5 Harmonic Series and p-Series.

✏️ Finding Terms in A Sequence: Example 2

{(1)nn!2n}1\{\frac{(-1)^n \cdot n!}{2^n}\}^\infty_1

This is the same process as the last example, but here, we can use some algebra to simplify an+1a_{n+1} bit!

a1=12a_1 = -\dfrac12
a2=12a_2 = \dfrac12
an=(1)nn!2na_n = \frac{(-1)^n \cdot n!}{2^n}
an+1=(1)n+1(n+1)!2n+1=(1)n(n+1)n!2(2n)=ann+12a_{n+1} = \frac{(-1)^{n+1}\cdot(n+1)!}{2^n+1} = -\frac{(-1)^n(n+1)n!}{2(2^n)} = -a_n\cdot\frac{n+1}{2}
(1)n(-1)^n

This is a special type of sequence called an alternating sequence. You can identify this by the (1)n(-1)^n in the sequence formula, and we will learn more about alternating sequences and sequences in the following study guide: 10.7 Alternating Series Test for Convergence.


🛑 The Limit of A Sequence

Like functions, sequences have limits too! These are found in much the same way that the limit of a function is found, but in this unit, we are only interested in finding out the limit as n approaches \infty. Also, note that all limit properties that hold for regular functions hold for sequences as well.

There is also a special theorem that holds for sequence limits as well which will be useful for. We will not be doing any examples specifically geared towards finding limits of sequences, but these will be used in other applications as well.

The first of these is finding out whether a sequence is convergent or divergent. The words convergent and divergent will show up a lot in this unit, so stay alert!

  1. Convergent Sequence: A sequence in which limnan\lim\limits_{n \to \infty} a_n exists and is finite.
  2. Divergent Sequence: A sequence in which limnan\lim\limits_{n \to \infty} a_n does not exist or is infinite.

🔁 Convergence and Divergence of Sequences

Determine whether the following sequences converge or diverge.

✏️ Sequences Example 1: Converging or Diverging?

{(1)nn}1\{\frac{(-1)^n}{n}\}^\infty_1
=limn(1)nn= \lim\limits_{n \to \infty} \frac{(-1)^n}{n}
=limn(1)n1n=0= \lim\limits_{n \to \infty} (-1)^n\cdot\dfrac1n = 0

Because the limit is finite, the sequence converges.

✏️ Sequences Example 2: Converging or Diverging?

{n2+1n}1\{\frac{n^2+1}{n}\}^\infty_1

Based on what we have learned about limits of rational functions, =limn n2 +1n== \lim\limits_{n \to \infty} \frac{\ n^2 \ +1}{n} = \infty

Thus, this sequence diverges.

✏️ Sequences Example 3: Converging or Diverging?

{(1)n}1\{(-1)^n\}^\infty_1

To find this limit, realize that this sequence is a piecewise function.

an={1n%2=01n%2=1a_n=\begin{cases}1& n\%2=0\\ -1& n\%2=1\\\end{cases}

As n approaches infinity, the sequence keeps oscillating between 1 and -1, thus the limit does not exist and the sequence diverges.


‼️ Some Notation and Terminology on Sequences

Before we move on to series, there is some terminology that we have to cover real quick when we talk about sequences.

  • Increasing Sequence: When an+1an>1\frac{a^{n+1}}{a_n} > 1 for all n, meaning that each term is greater than the last.
  • Decreasing Sequence: When an+1an<1\frac{a^{n+1}}{a_n} < 1 for all n, meaning that each term is less than the last.
  • Monotonic Sequence: A sequence that is either increasing or decreasing.
  • Bounded Above: When there exists an upper bound to the sequence, meaning that the terms do not go above a certain amount.
  • Bounded Below: When there exists a lower bound to the sequence, meaning that the terms do not go below a certain amount.
  • Bounded: When the sequence is both bounded above or below.

These definitions lead to a theorem about sequence convergence.

Sequence Convergence Theorem

If a sequence is both bounded and monotonic, then the sequence converges. Note: It does not mean that if the sequence does not meet both conditions, then it diverges.

Now that we understand sequences, let’s start talking about series!


😵‍💫 What is a Series?

A series is just a sum of the terms in a sequence. The definition is as simple as that! A series can be written like this:

sn=i=1nais_n = \sum_{i=1}^{n} a_i

where sns_n is the series of the sequence aia_i, summing up all the terms from the 1st term to the nthn^{th} term, inclusive. The nth**n^{th} partial sum** is the value of the summation of the 1st through the nthn^{th} terms. An infinite series is a series where n=\infty, or more specifically.

s=limni=1nais_\infty = \lim\limits_{n \to \infty} \sum_{i=1}^{n} a_i

We usually cannot find the sum of an infinite series, but in the following example, we will be able to find the sum of this series.

🤓 Finding Partial Sums

Find the partial sums s1,s2,s3,sn,s_1, s_2,s_3, s_n, and ss_∞of the series with an=1n2+na_n=\frac {1}{n^2 +n} .

To start, let’s find the values ofa1,a2,a_1 ,a_2 , and a3.a_3.

a1=12a_1=\frac {1}{2}

a2=16a_2=\frac {1}{6}

a3=112a_3=\frac {1}{12}

From this, we can figure out the first three partial sums!

s1=a1=12s_1 = a_1 = \dfrac12
s2=a1+a2=46=23s_2 = a_1 + a_2 = \dfrac46 = \dfrac23
s3=a1+a2+a3=912=34s_3 = a_1+a_2+a_3 = \dfrac{9}{12} = \dfrac34

Before we continue to find sns_n and ss_∞, we’re going to do a partial fraction decomposition of the sequence formula. It is unlikely that you will have to do this on the AP test, but this is to show that in some cases, ss_∞ can be found.

an=1n2+n=1n(n+1)=1n1n+1a_n = \frac{1}{n^2+n} = \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}

Now that we have this, let’s find sns_n and ss_∞!

sn=1112+1213+13+...+1n11n+1n1n+1s_n = \frac{1}{1} - \dfrac12 + \dfrac12 - \dfrac13 + \dfrac13 +...+ \dfrac{1}{n-1} - \dfrac{1}{n} + \dfrac{1}{n} - \dfrac{1}{n+1}
=11n+1= 1 - \frac{1}{n+1}

As you can see here, the middle terms in the expansion all cancel out, leaving us with only the first and last terms, allowing us to cancel out the other terms. Any series that acts this way is called a telescoping series, which is kind of like a collapsible telescope in which the middle collapses and the ends remain. But finally, it’s time for us to find ss_∞!

s=limnsns_\infty = \lim\limits_{n \to \infty} s_n
=limn11n+1= \lim\limits_{n \to \infty} 1-\frac{1}{n+1}
=1=1

🧐 Convergence and Divergence of Series

Like sequences, series can also converge or diverge. We will list their definitions below.

  1. Convergent Series: A series in which ss_\infty exists and is finite.
  2. Divergent Series: A series in which ss_\infty does not exist or is infinite.

Since the series we just did has a finite value for the infinite partial sum, the series converges. In the rest of the first part of the unit, we will find a way to determine whether a series is convergent or divergent, so don’t worry if you don’t know this yet!

Here are some properties of convergent series that will be helpful throughout the unit! Given that sns_n and tnt_n are convergent series and cc is a constant, the following properties apply.

  1. The series rn=csnr_n = c\cdot s_n is convergent
  2. The series qn=sn±tnq_n = s_n \pm t_n is also convergent

✏️ Partial Sums: Practice Problems

Using the above examples, do these next two on your own! Try them before looking at the answers.

For each sequence an,a_n , find an+1a_{n+1} in terms of ana_n if possible and then solve for the partial sums s1,s2,s_1,s_2 , and s3s_3.

1. an=1en1. \ a_n = \frac{1}{e^n}
2. an=n(n+2)!2.\ a_n = \frac{n}{(n+2)!}

✅ Solution to Partial Sums Example 1

No peeking until you’ve given the question a try!

an=1ena_n = \frac{1}{e^n}
an+1=e(n+1)a_{n+1}= e^{-(n+1)}
s1=e1s_1=e^{-1}
s2=e1+e2s_2=e^{-1}+e^{-2}
s3=e1+e2+e3s_3=e^{-1}+e^{-2}+e^{-3}

You made it through this question!

✅ Solution to Partial Sums Example 2

Last but not least…

an=n(n+2)!a_n = \frac{n}{(n+2)!}
an+1=(n+1)(n+3)!a_{n+1}=\frac{(n+1)}{(n+3)!}
s1=16s_1=\frac{1}{6}
s2=14s_2=\frac{1}{4}
s3=1140s_3=\frac{11}{40}

Great work!


⭐️ Conclusion

If you are overwhelmed, don’t worry! The more you do in unit 10, the easier these problems will get. The entirety of this unit is building your toolbox to be able to solve these problems. Continue on your journey to be a calculus wizard 🧙

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10.1 Defining Convergent and Divergent Infinite Series

2 min readjune 18, 2024

Avanish Gupta

Avanish Gupta

Avanish Gupta

Avanish Gupta

10.1 Defining Convergent and Divergent Infinite Series

Welcome to unit 10 of AP Calculus BC! This is the last one. 🥳

This first, longer part of the guide concerns the behavior of numerical series and whether they have an actual sum as the number of terms approaches infinity. There are many tests and methods that we can use to answer this question, each working on specific cases. However, before we can answer this, we need to first talk about what sequences and series are.

Before we start talking about series, we need to talk about sequences and some terminology about sequences.

🤔 What is a Sequence?

A sequence is just a list of terms related by a common pattern to each other.

Here is how we represent sequences:

{an}1\{a_n\}^\infty_1

📝 Terms in a Sequence

List a1,a2,a3,ana_1, a_2, a_3, a_n, and an+1a_{n + 1} for the following sequences. We can go through this together!

✏️ Finding Terms in A Sequence: Example 1

{1n}1\{\dfrac1n\}^\infty_1

This is nothing more than just plugging in n = 1, 2, 3, n, and n+1 into ana_n.

{1,12,13,...,1n,1n+1,...}\{1, \dfrac12, \dfrac13,...,\frac{1}{n}, \frac{1}{n+1}, ...\}

We use the ... between the 3rd and the nthn^{th} term and after the (n+1)th(n+1)^{th} term because there are an indefinite number of terms in these intervals.

This is a special sequence called the harmonic sequence. We will learn more about the following study guide later in this unit: 10.5 Harmonic Series and p-Series.

✏️ Finding Terms in A Sequence: Example 2

{(1)nn!2n}1\{\frac{(-1)^n \cdot n!}{2^n}\}^\infty_1

This is the same process as the last example, but here, we can use some algebra to simplify an+1a_{n+1} bit!

a1=12a_1 = -\dfrac12
a2=12a_2 = \dfrac12
an=(1)nn!2na_n = \frac{(-1)^n \cdot n!}{2^n}
an+1=(1)n+1(n+1)!2n+1=(1)n(n+1)n!2(2n)=ann+12a_{n+1} = \frac{(-1)^{n+1}\cdot(n+1)!}{2^n+1} = -\frac{(-1)^n(n+1)n!}{2(2^n)} = -a_n\cdot\frac{n+1}{2}
(1)n(-1)^n

This is a special type of sequence called an alternating sequence. You can identify this by the (1)n(-1)^n in the sequence formula, and we will learn more about alternating sequences and sequences in the following study guide: 10.7 Alternating Series Test for Convergence.


🛑 The Limit of A Sequence

Like functions, sequences have limits too! These are found in much the same way that the limit of a function is found, but in this unit, we are only interested in finding out the limit as n approaches \infty. Also, note that all limit properties that hold for regular functions hold for sequences as well.

There is also a special theorem that holds for sequence limits as well which will be useful for. We will not be doing any examples specifically geared towards finding limits of sequences, but these will be used in other applications as well.

The first of these is finding out whether a sequence is convergent or divergent. The words convergent and divergent will show up a lot in this unit, so stay alert!

  1. Convergent Sequence: A sequence in which limnan\lim\limits_{n \to \infty} a_n exists and is finite.
  2. Divergent Sequence: A sequence in which limnan\lim\limits_{n \to \infty} a_n does not exist or is infinite.

🔁 Convergence and Divergence of Sequences

Determine whether the following sequences converge or diverge.

✏️ Sequences Example 1: Converging or Diverging?

{(1)nn}1\{\frac{(-1)^n}{n}\}^\infty_1
=limn(1)nn= \lim\limits_{n \to \infty} \frac{(-1)^n}{n}
=limn(1)n1n=0= \lim\limits_{n \to \infty} (-1)^n\cdot\dfrac1n = 0

Because the limit is finite, the sequence converges.

✏️ Sequences Example 2: Converging or Diverging?

{n2+1n}1\{\frac{n^2+1}{n}\}^\infty_1

Based on what we have learned about limits of rational functions, =limn n2 +1n== \lim\limits_{n \to \infty} \frac{\ n^2 \ +1}{n} = \infty

Thus, this sequence diverges.

✏️ Sequences Example 3: Converging or Diverging?

{(1)n}1\{(-1)^n\}^\infty_1

To find this limit, realize that this sequence is a piecewise function.

an={1n%2=01n%2=1a_n=\begin{cases}1& n\%2=0\\ -1& n\%2=1\\\end{cases}

As n approaches infinity, the sequence keeps oscillating between 1 and -1, thus the limit does not exist and the sequence diverges.


‼️ Some Notation and Terminology on Sequences

Before we move on to series, there is some terminology that we have to cover real quick when we talk about sequences.

  • Increasing Sequence: When an+1an>1\frac{a^{n+1}}{a_n} > 1 for all n, meaning that each term is greater than the last.
  • Decreasing Sequence: When an+1an<1\frac{a^{n+1}}{a_n} < 1 for all n, meaning that each term is less than the last.
  • Monotonic Sequence: A sequence that is either increasing or decreasing.
  • Bounded Above: When there exists an upper bound to the sequence, meaning that the terms do not go above a certain amount.
  • Bounded Below: When there exists a lower bound to the sequence, meaning that the terms do not go below a certain amount.
  • Bounded: When the sequence is both bounded above or below.

These definitions lead to a theorem about sequence convergence.

Sequence Convergence Theorem

If a sequence is both bounded and monotonic, then the sequence converges. Note: It does not mean that if the sequence does not meet both conditions, then it diverges.

Now that we understand sequences, let’s start talking about series!


😵‍💫 What is a Series?

A series is just a sum of the terms in a sequence. The definition is as simple as that! A series can be written like this:

sn=i=1nais_n = \sum_{i=1}^{n} a_i

where sns_n is the series of the sequence aia_i, summing up all the terms from the 1st term to the nthn^{th} term, inclusive. The nth**n^{th} partial sum** is the value of the summation of the 1st through the nthn^{th} terms. An infinite series is a series where n=\infty, or more specifically.

s=limni=1nais_\infty = \lim\limits_{n \to \infty} \sum_{i=1}^{n} a_i

We usually cannot find the sum of an infinite series, but in the following example, we will be able to find the sum of this series.

🤓 Finding Partial Sums

Find the partial sums s1,s2,s3,sn,s_1, s_2,s_3, s_n, and ss_∞of the series with an=1n2+na_n=\frac {1}{n^2 +n} .

To start, let’s find the values ofa1,a2,a_1 ,a_2 , and a3.a_3.

a1=12a_1=\frac {1}{2}

a2=16a_2=\frac {1}{6}

a3=112a_3=\frac {1}{12}

From this, we can figure out the first three partial sums!

s1=a1=12s_1 = a_1 = \dfrac12
s2=a1+a2=46=23s_2 = a_1 + a_2 = \dfrac46 = \dfrac23
s3=a1+a2+a3=912=34s_3 = a_1+a_2+a_3 = \dfrac{9}{12} = \dfrac34

Before we continue to find sns_n and ss_∞, we’re going to do a partial fraction decomposition of the sequence formula. It is unlikely that you will have to do this on the AP test, but this is to show that in some cases, ss_∞ can be found.

an=1n2+n=1n(n+1)=1n1n+1a_n = \frac{1}{n^2+n} = \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}

Now that we have this, let’s find sns_n and ss_∞!

sn=1112+1213+13+...+1n11n+1n1n+1s_n = \frac{1}{1} - \dfrac12 + \dfrac12 - \dfrac13 + \dfrac13 +...+ \dfrac{1}{n-1} - \dfrac{1}{n} + \dfrac{1}{n} - \dfrac{1}{n+1}
=11n+1= 1 - \frac{1}{n+1}

As you can see here, the middle terms in the expansion all cancel out, leaving us with only the first and last terms, allowing us to cancel out the other terms. Any series that acts this way is called a telescoping series, which is kind of like a collapsible telescope in which the middle collapses and the ends remain. But finally, it’s time for us to find ss_∞!

s=limnsns_\infty = \lim\limits_{n \to \infty} s_n
=limn11n+1= \lim\limits_{n \to \infty} 1-\frac{1}{n+1}
=1=1

🧐 Convergence and Divergence of Series

Like sequences, series can also converge or diverge. We will list their definitions below.

  1. Convergent Series: A series in which ss_\infty exists and is finite.
  2. Divergent Series: A series in which ss_\infty does not exist or is infinite.

Since the series we just did has a finite value for the infinite partial sum, the series converges. In the rest of the first part of the unit, we will find a way to determine whether a series is convergent or divergent, so don’t worry if you don’t know this yet!

Here are some properties of convergent series that will be helpful throughout the unit! Given that sns_n and tnt_n are convergent series and cc is a constant, the following properties apply.

  1. The series rn=csnr_n = c\cdot s_n is convergent
  2. The series qn=sn±tnq_n = s_n \pm t_n is also convergent

✏️ Partial Sums: Practice Problems

Using the above examples, do these next two on your own! Try them before looking at the answers.

For each sequence an,a_n , find an+1a_{n+1} in terms of ana_n if possible and then solve for the partial sums s1,s2,s_1,s_2 , and s3s_3.

1. an=1en1. \ a_n = \frac{1}{e^n}
2. an=n(n+2)!2.\ a_n = \frac{n}{(n+2)!}

✅ Solution to Partial Sums Example 1

No peeking until you’ve given the question a try!

an=1ena_n = \frac{1}{e^n}
an+1=e(n+1)a_{n+1}= e^{-(n+1)}
s1=e1s_1=e^{-1}
s2=e1+e2s_2=e^{-1}+e^{-2}
s3=e1+e2+e3s_3=e^{-1}+e^{-2}+e^{-3}

You made it through this question!

✅ Solution to Partial Sums Example 2

Last but not least…

an=n(n+2)!a_n = \frac{n}{(n+2)!}
an+1=(n+1)(n+3)!a_{n+1}=\frac{(n+1)}{(n+3)!}
s1=16s_1=\frac{1}{6}
s2=14s_2=\frac{1}{4}
s3=1140s_3=\frac{11}{40}

Great work!


⭐️ Conclusion

If you are overwhelmed, don’t worry! The more you do in unit 10, the easier these problems will get. The entirety of this unit is building your toolbox to be able to solve these problems. Continue on your journey to be a calculus wizard 🧙