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8.1: Sequences

Difficulty Level: At Grade Created by: CK-12
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Learning Objectives

  • Demonstrate an understanding of sequences and their terms
  • Determine if the limit of a sequence exists and, if it exists, find the limit
  • Apply rules, theorems, and Picard’s method to compute the limits of sequences

Sequences (rules, terms, indices)

The alphabet, the names in a phone book, the numbered instructions of a model airplane kit, and the schedule in the local television guide are examples of sequences people may use. These examples are all sets of ordered items. In mathematics, a sequence is a list of numbers. You can make finite sequences, such as \begin{align*}2, 4, 6, 8\end{align*}. These sequences end. You can also make infinite sequences, such as \begin{align*}3, 5, 7, 9, \ldots,\end{align*} which do not end but continue on as indicated by the three dots. In this chapter the word sequence refers to an infinite sequence.

Each term in a sequence is defined by its place of order in the list. Consider the sequence \begin{align*}3, 5, 7, 9, \ldots\end{align*} The first term is \begin{align*}3\end{align*} because it belongs to place \begin{align*}1\end{align*} of the sequence. The second term is \begin{align*}5\end{align*} because it belongs to the second place of the sequence. Likewise, The third term is \begin{align*}7\end{align*} because it is in the third place. Notice that there is a natural relationship between the counting numbers, or the positive integers, and the terms of the sequence. This leads us to the definition of a sequence.


A sequence is a function from the domain of the set of counting numbers, or positive integers, to the range which consists of the members of a sequence.

A sequence can be denoted by \begin{align*}\left \{a_n \right \}\end{align*} or by \begin{align*}a_1, a_2, a_3, a_4, \ldots, a_n,\ldots\end{align*}

The numbers \begin{align*}a_1, a_2, a_3, a_4, \ldots, a_n,\ldots\end{align*} that belong to a sequence are called terms of the sequence. Each subscript of \begin{align*}1, 2, 3, \ldots\end{align*} on the terms \begin{align*}a_1, a_2, a_3, a_4,\ldots\end{align*} refers to the place of the terms in the sequence, or the index. The subscripts are called the indices of the terms. We assume that \begin{align*}n = 1, 2, 3,\ldots,\end{align*} unless otherwise noted.

Instead of listing the elements of a sequence, we can define a sequence by a rule, or formula, in terms of the indices.

Example 1

The formula \begin{align*}a_n = \frac{1} {n}\end{align*} is a rule for a sequence.

We can generate the terms for this rule as follows:

\begin{align*}& n&& 1&& 2&& 3&& 4 && \ldots \\ & a_n = \frac{1} {n}&& \frac{1} {1} = 1&& \frac{1} {2}&& \frac{1} {3}&& \frac{1} {4} && \ldots\end{align*}

Example 2

Consider the sequence rule \begin{align*}a_n = \frac{n^2} {n + 1}\end{align*}.

The terms of the sequence are:

\begin{align*}& n && 1 && 2 && 3 && 4 && \ldots \\ & a_n = \frac{n^2} {n + 1} && \frac{1^2} {1 + 1} = \frac{1} {2} && \frac{2^2} {2 + 1} = \frac{4} {3} && \frac{9} {4} && \frac{16} {5} && \ldots\end{align*}

You can also find the rule for a sequence.

Example 3

Find the rule for the sequence below.

\begin{align*}& n && 1 && 2 && 3 && 4 && \ldots \\ & a_n = ? && \frac{1} {2} && -\frac{2} {3} && \frac{3} {4} && -\frac{4} {5} && \ldots\end{align*}

Look at each term in terms of its index. The numerator of each term matches the index. The denominator is one more than the index. So far, we can write the formula \begin{align*}a_n\end{align*} as \begin{align*}\frac{n} {n + 1}\end{align*}. However, we are not done. Notice that each even-indexed term has a negative sign. This means that all of terms of the sequence have a power of \begin{align*}-1\end{align*}. The powers of \begin{align*}-1\end{align*} alternate between odd and even. Usually, alternating powers of \begin{align*}-1\end{align*} can be denote by \begin{align*}(-1)^n\end{align*} or \begin{align*}(-1)^{n + 1}\end{align*}. Since the terms are negative for even indices, we use \begin{align*}(-1)^{n + 1}\end{align*}. Thus, the rule for the sequence is \begin{align*}a_n = \frac{(-1)^{n+1} n} {n + 1}\end{align*}. You can check the rule by finding the first few terms of the sequence \begin{align*}a_n = \frac{(-1)^{n+1} n} {n + 1}\end{align*}.

Limit of a Sequence

We are interested in the behavior of the sequence as the value of \begin{align*}n\end{align*} gets very large. Many times a sequence will get closer to a certain number, or limit, as \begin{align*}n\end{align*} gets large. Finding the limit of a sequence is very similar to finding the limit of a function. Let’s look at some graphs of sequences.

Example 4

Find the limit of the sequence \begin{align*}\left \{ \frac{1} {2n + 1}\right \}\end{align*} as \begin{align*}n\end{align*} goes to infinity.


We can graph the corresponding function \begin{align*}y = \frac{1} {2n + 1}\end{align*} for \begin{align*}n = 1, 2, 3,\ldots.\end{align*} The graph of is similar to the continuous function \begin{align*}y = \frac{1} {2x + 1}\end{align*} for the domain of \begin{align*}x \ge 1\end{align*}.

To determine the limit, we look at the trend or behavior of the graph of sequence as \begin{align*}n\end{align*} gets larger or travels out to positive infinity. This means we look at the points of sequence that correspond to the far right end of the horizontal axis in the figure on the right. We see that the points of the sequence are getting closer to the horizontal axis, \begin{align*}y = 0\end{align*}. Thus, the limit of the sequence \begin{align*}\left \{ \frac{1} {2n + 1}\right \}\end{align*} is as \begin{align*}n\end{align*} tends to infinity. We write: \begin{align*}\lim_{n \to +\infty} \frac{1} {2n + 1} = 0 .\end{align*}

Here is the precise definition of the limit of a sequence.

Limit of a Sequence

The limit of a sequence \begin{align*}a_n\end{align*} is the number \begin{align*}L\end{align*} if for each \begin{align*}\varepsilon > 0\end{align*}, there exists an integer \begin{align*}N\end{align*} such that \begin{align*}|\ a_n - L\ | < \varepsilon\end{align*} for all \begin{align*}n > N.\end{align*}

Recall that \begin{align*}|\ a_n - L\ | < \varepsilon\end{align*} means the values of \begin{align*}a_n\end{align*} such that \begin{align*}L - \varepsilon < a_n < L + \varepsilon.\end{align*}

What does the definition of the limit of a sequence mean? Here is another example.

Example 5

Look at Figure 3.

Figure 3

Figure 3 shows the graph of the sequence \begin{align*}\left \{ \frac{l_n \ (n)} {n}\right \}\end{align*}. Notice that from \begin{align*}N\end{align*} on, the terms of \begin{align*}\frac{l_n\ n} {n}\end{align*} are between \begin{align*}L - \varepsilon\end{align*} and \begin{align*}L + \varepsilon\end{align*}. In other words, for this value of \begin{align*}\varepsilon,\end{align*} there is a value \begin{align*}N\end{align*} such that all terms of \begin{align*}a_n\end{align*} are in the interval from \begin{align*}L - \varepsilon\end{align*} and \begin{align*}L + \varepsilon\end{align*}. Thus, \begin{align*}\lim_{n \to +\infty} \frac{l_n\ (n)} {n} = 0\end{align*}.

Not every sequence has a limit.

Example 6

Here is a graph of the sequence \begin{align*}\left \{n + 1 \right \}\end{align*}.

Figure 4

Consider the sequence \begin{align*}\left \{n + 1 \right \}\end{align*} in Figure 4. As \begin{align*}n\end{align*} gets larger and goes to infinity, the terms of \begin{align*}a_n = n + 1\end{align*} become larger and larger. The sequence \begin{align*}\left \{n + 1 \right \}\end{align*} does not have a limit. We write \begin{align*}\lim_{n \to +\infty} (n + 1) = +\infty.\end{align*}

Convergence and Divergence

We say that a sequence \begin{align*}\left \{a_n \right \}\end{align*} converges to a limit \begin{align*}L\end{align*} if sequence has a finite limit \begin{align*}L\end{align*}. The sequence has convergence. We describe the sequence as convergent. Likewise, a sequence \begin{align*}\left \{a_n \right \}\end{align*} diverges to a limit \begin{align*}L\end{align*} if sequence does not have a finite limit. The sequence has divergence and we describe the sequence as divergent.

Example 7

The sequence \begin{align*}\left \{\ln\ (n) \right \}\end{align*} grows without bound as \begin{align*}n\end{align*} approaches infinity. Note that the related function \begin{align*}y = \ln(x)\end{align*} grows without bound. The sequence is divergent because it does not have a finite limit. We write \begin{align*}\lim_{n \to +\infty} \ln\ (n) = +\infty\end{align*}.

Example 8

The sequence \begin{align*}\left \{ 4 -\frac{8} {n}\right \}\end{align*} converges to the limit \begin{align*}L = 4\end{align*} and hence is convergent. If you graph the function \begin{align*}y = 4 - \frac{8} {n}\end{align*} for \begin{align*}n = 1, 2, 3,\ldots,\end{align*} you will see that the graph approaches \begin{align*}4\end{align*} as \begin{align*}n\end{align*} gets larger. Algebraically, as \begin{align*}n\end{align*} goes to infinity, the term \begin{align*}-\frac{8} {n}\end{align*} gets smaller and tends to while \begin{align*}4\end{align*} stays constant. We write \begin{align*}\lim_{n \to +\infty} \left (4 - \frac{8} {n}\right ) = 4\end{align*}.

Example 9

Does the sequence \begin{align*}s_n\end{align*} with terms \begin{align*}1, -1, 1, -1, 1, -1, \ldots.\end{align*} have a limit?


This sequence oscillates, or goes back and forth, between the values \begin{align*}1\end{align*} and \begin{align*}-1\end{align*}. The sequence does not get closer to \begin{align*}1\end{align*} or \begin{align*}-1\end{align*} as \begin{align*}n\end{align*} gets larger. We say that the sequence does not have a limit, or \begin{align*}\lim_{n \to +\infty} s_n\end{align*} does not exist.

Note: Each sequence’s limit falls under only one of the four possible cases:

  1. A limit exists and the limit is \begin{align*}L\end{align*}: \begin{align*}\lim_{n \to +\infty} s_n = L\end{align*}.
  2. There is no limit: \begin{align*}\lim_{n \to +\infty} s_n\end{align*} does not exist.
  3. The limit grows without bound in the positive direction and is divergent: \begin{align*}\lim_{n \to +\infty} s_n = +\infty\end{align*}.
  4. The limit grows without bound in the negative direction and is divergent: \begin{align*}\lim_{n \to +\infty}s_n = -\infty.\end{align*}

If a sequence has a finite limit, then it only has one value for that limit.


If a sequence is convergent, then its limit is unique.

Keep in mind that being divergent is not the same as not having a limit.

L’Hôpital’s Rule

Realistically, we cannot graph every sequence to determine if it has a finite limit and the value of that limit. Nor can we make an algebraic argument for the limit for every possible sequence. Just as there are indeterminate forms when finding limits of functions, there are indeterminate forms of sequences, such as \begin{align*}\frac{0} {0}, \frac{\infty} {\infty}, 0 + \infty\end{align*}. To find the limit of such sequences, we can apply L’Hôpital’s rule.

Example 10

Find \begin{align*}\lim_{n \to +\infty} \frac{ \ln\ (n)} {n}\end{align*}.


We solved this limit by using a graph in Example 5. Let’s solve this problem using L’Hôpital’s rule. The numerator is \begin{align*}\ln\ (n)\end{align*} and the denominator is \begin{align*}n\end{align*}. Both functions \begin{align*}y = \ln\ (n)\end{align*} and \begin{align*}y = n\end{align*} do not have limits. So, the sequence \begin{align*}\left \{ \frac{\ln\ (n)} {n}\right \}\end{align*} is of the indeterminate form \begin{align*}\frac{\infty} {\infty}\end{align*}. Since the functions \begin{align*}y = \ln\ (n)\end{align*} and \begin{align*}y = n\end{align*} are not differentiable, we apply L’Hôpital’s rule to the corresponding problem, \begin{align*}\lim_{x \to +\infty} \frac{\ln\ (x)} {x}\end{align*}, first. Taking the first derivative of the numerator and denominator of \begin{align*}y = \frac{\ln\ (x)} {x}\end{align*}, we find \begin{align*}\lim_{x \to +\infty} \frac{\ln\ (x)} {x} = \lim_{x \to +\infty} \frac{\frac{1} {x}} {1} = 0\end{align*}. Thus, \begin{align*}\lim_{n \to +\infty} \frac{\ln\ (n)} {n} = 0\end{align*} because the points of \begin{align*}y = \frac{\ln\ (n)} {n}\end{align*} are a subset of the points of the function \begin{align*}y = \frac{\ln\ (x)} {x}\end{align*} as \begin{align*}x\end{align*} approaches infinity. We also confirmed the limit of the sequence with its graph in Example \begin{align*}5\end{align*}.

Rules, Sandwich/Squeeze

Properties of function limits are also used with limits of sequences.

Theorem (Rules)

Let \begin{align*}\left \{a_n \right \}\end{align*} and \begin{align*}\left \{b_n \right \}\end{align*} be sequences such that \begin{align*}\lim_{n \to +\infty} a_n = L_1\end{align*} and \begin{align*}\lim_{n \to +\infty} b_n = L_2\end{align*}.

Let \begin{align*}c\end{align*} be any constant. Then the following statements are true:

1. \begin{align*}\lim_{n \to +\infty} c = c\end{align*}

The limit of a constant is the same constant.

2. \begin{align*}\lim_{n \to +\infty} c \times a_n = c \times \lim_{n \to +\infty} a_n = cL_1\end{align*}

The limit of a constant times a sequence is the same as the constant times the limit of the sequence.

3. \begin{align*}\lim_{n \to +\infty} (a_n + b_n) = \lim_{n \to +\infty} a_n + \lim_{n \to +\infty} = L_1 + L_2\end{align*}

The limit of a sum of sequences is the same as the sum of the limits of the sequences.

4. \begin{align*}\lim_{n \to +\infty} (a_n \times b_n) = \lim_{n \to +\infty} a_n \times \lim_{n \to +\infty} b_n = L_1 L_2\end{align*}

The limit of the product of sequences is the same as the product of the limits of the sequences.

5. If \begin{align*}L_2 \ne 0,\end{align*} then \begin{align*}\lim_{n \to +\infty} \left (\frac{a_n} {b_n}\right ) = \frac{\lim_{n \to +\infty}a_n} {\lim_{n \to +\infty} b_n} = \frac{L_1} {L_2}\end{align*}.

The limit of the quotient of two sequences is the same as the quotient of the limits of the sequences.

Let’s apply these rules to help us find limits.

Example 11

Find \begin{align*}\lim_{n \to +\infty} \frac{7n} {9n + 5}\end{align*}.


We could use L’Hôpital’s rule or we could use some of the rules in the preceding theorem. Let’s use the rules in the theorem. Divide both the numerator and denominator by the highest power of \begin{align*}n\end{align*} in the expression and using rules from the theorem, we find the limit:

\begin{align*}\lim_{n \to +\infty} \frac{7n} {9n + 5} & = \lim_{n \to +\infty} \frac{\frac{7n} {n}} {\frac{9n} {n} + \frac{5} {n}}\ \text{Dividing both numerator and denominator by}\ n\\ & = \lim_{n \to +\infty} \frac{7} {\left (9 + \frac{5} {n}\right )}\ \text{Simplifying} \\ & = \frac{\lim_{n \to +\infty} 7} {\lim_{n \to +\infty} \left (9 + \frac{5} {n}\right )}\ \text{Applying the division rule for limits.}\\ & = \frac{\lim_{n \to +\infty} 7} {\lim_{n \to +\infty} 9 + \lim_{n \to +\infty} \frac{5} {n}} \ \text{Applying the rule for the limit of a sum to the denominator} \\ & = \frac{7} {9 + 0} = \frac{7} {9}\ \text{Evaluating the limits}\end{align*}

Example 12

Find \begin{align*}\lim_{n \to +\infty} \left (\frac{11} {n} - \frac{8} {n^2}\right )\end{align*}.


\begin{align*}\lim_{n \to +\infty} \left (\frac{11} {n} - \frac{8} {n^2}\right ) & = \lim_{n \to +\infty} \frac{11} {n} - \lim_{n \to +\infty} \frac{8} {n^2}\ \text{Applying the rule for the difference of two limits} \\ & = 11 \lim_{n \to +\infty} \frac{1} {n} - 8 \lim_{n \to +\infty} \frac{1} {n^2}\ \text{Applying the rule for the limit of c times a limit} \\ & = 11 \times 0 - 8 \times 0 = 0\ \text{Evaluating the limits}\end{align*}

As with limits of functions, there is a Sandwich/Squeeze Theorem for the limits of sequences.

Sandwich/Squeeze Theorem

Let \begin{align*}\left \{a_n \right \}, \left \{b_n \right \}\end{align*} and \begin{align*}\left \{c_n \right \}\end{align*} be sequences. Let \begin{align*}N\end{align*} be a positive integer.

Suppose \begin{align*}c_n\end{align*} is a sequence such that \begin{align*}a_n \le c_n \le b_n\end{align*} for all \begin{align*}n \ge N\end{align*}. Suppose also that

\begin{align*}\lim a_n = \lim b_n = L\end{align*}. Then \begin{align*}\lim c_n = L\end{align*}.

You can see how the name of the theorem makes sense from the statement. After a certain point in the sequences, the terms of a sequence \begin{align*}c_n\end{align*} are sandwiched or squeezed between the terms of two convergent sequences with the same limit. Then the limit of the sequence \begin{align*}c_n\end{align*} is squeezed to become the same as the limit of the two convergent sequences. Let’s look at an example.

Example 13

Prove \begin{align*}\lim_{n \to +\infty} \frac{8^n} {n!} = 0\end{align*}.


Recall that \begin{align*}n!\end{align*} is read as “n factorial” and is written as \begin{align*}n! = n \times (n - 1) \times (n - 2)\times \ldots \times 1.\end{align*}

We want to apply the Sandwich theorem by squeezing the sequence \begin{align*}\frac{8^n} {n!}\end{align*} between two sequences that converge to the same limit.

First, we know that \begin{align*}0 \le \frac{8^n} {n!}\end{align*}. Now we want to find a sequence whose terms greater than or equal to the terms of the sequence \begin{align*}\frac{8}{7}\end{align*} for some \begin{align*}n\end{align*}.

We can write

\begin{align*}\frac{8^n} {n!} & = \frac{8 \times 8 \times 8 \ldots \times 8} {n \times (n - 1) \times (n - 2) \times \ldots 1} \\ & = \frac{8} {n} \times \frac{8} {n - 1} \times \ldots \times \frac{8} {2} \times \frac{8} {1} \\ & = \left (\frac{8} {n}\right ) \left (\frac{8} {n - 1} \times \ldots \times \frac{8} {9} \times \frac{8} {8}\right ) \left (\frac{8} {7} \times \frac{8} {6} \times \frac{8} {5} \times \ldots \times \frac{8} {1} \right )\end{align*}

Since each factor in the product \begin{align*}\frac{8} {n - 1} \times \ldots \times \frac{8} {9} \times \frac{8} {8}\end{align*} is less than or equal to 1, then the product \begin{align*}\frac{8} {n - 1} \times \ldots \times \frac{8} {9} \times \frac{8} {8} \le 1\end{align*}. Then we make an inequality:

\begin{align*}\left (\frac{8} {n}\right ) \left (\frac{8} {n - 1} \times \ldots \times \frac{8} {9} \times \frac{8} {8}\right ) \left (\frac{8} {7} \times \frac{8} {6} \times \frac{8} {5} \times \ldots \times \frac{8} {1} \right ) & \le \left (\frac{8} {7}\right ) (1) \left (\frac{8} {7} \times \frac{8} {6} \times \frac{8} {5} \times \ldots \times \frac{8} {1} \right ) \\ & = \left (\frac{8} {7} \right ) \left (\frac{8} {7} \times \frac{8} {6} \times \frac{8} {5} \times \ldots \times \frac{8} {1} \right ) \\ & = \left (\frac{8} {n} \right ) \left (\frac{8^7} {7!}\right )\end{align*}

Thus, \begin{align*}\lim_{n \to +\infty} 0 \le \lim_{n \to +\infty} \frac{8^n} {n!} \le \lim_{n \to +\infty} \left (\frac{8} {n}\right ) \left (\frac{8^7} {7!}\right )\end{align*}. By using the Rules Theorem, we have \begin{align*}\lim_{n \to +\infty} 0 = 0\end{align*} and \begin{align*}\lim_{n \to +\infty} \left (\frac{8} {n}\right ) \left (\frac{8^7} {7!}\right ) = \left (\frac{8^7} {7!}\right ) \lim_{n \to +\infty} \frac{8} {n} = \left (\frac{8^7} {7!}\right ) \times 0 = 0\end{align*}. Thus, \begin{align*}0 \le \lim_{n \to +\infty} \frac{8^n} {n!} \le 0\end{align*}. By the Sandwich/Squeeze Theorem, \begin{align*}\lim_{n \to +\infty} \frac{8^n} {n!} = 0\end{align*}.

Picard’s Method

The following method appeared in 1891 by Emile Picard, a famous French mathematician. It is a method for solving initial value problems in differential equations that produces a sequence of functions which converge to the solution. Start with the initial value problem:

\begin{align*}y' = f (x, y)\end{align*} with \begin{align*}y(x_0) = y_0\end{align*}

If \begin{align*}f (x, y)\end{align*} and \begin{align*}f_x (x, y)\end{align*} are both continuous then a unique solution to the initial value problem exists by Picard’s theory. Now if \begin{align*}y(x)\end{align*} is the solution to the given problem, then a reformulation of the differential equation is possible:

\begin{align*}\int^x_{x_0} y'(t) dt = \int_{x_0}^x f(t, y(t)) dt\end{align*}

Now the Fundamental Theorem of Calculus is utilized to integrate the left hand side of the problem and upon isolating , the following result is obtained:

\begin{align*}y(x) = y_0 + \int^x_{x_0} f(t, y(t)) dt\end{align*}

The equation above is the starting point for the Picard iteration because it will be used to build the sequence of functions which will describe the actual solution to the initial value problem. The Picard sequence of functions is calculated as follows:

Step 1 - Define \begin{align*}Y_0(x) = y_0\end{align*}

Step 2 - Substitute \begin{align*}Y_0(t) = y_0\end{align*} for \begin{align*}y(t)\end{align*} in \begin{align*}f(t, y(t))\end{align*}:

\begin{align*}Y_1(x) & = y_0 + \int_{x_0}^x f(t, Y_0 (t)) dt\\ Y_1(x) & = y_0 + \int_{x_0}^x f(t, Y_0 ) dt\end{align*}

Step 3 - Repeat step \begin{align*}2\end{align*} with \begin{align*}Y_1 (t)\end{align*} for \begin{align*}y(t)\end{align*} :

\begin{align*}Y_2(x) = y_0 + \int_{x_0}^x f(t, Y_1(t)) dt\end{align*}

The substitution process is repeated \begin{align*}n\end{align*} times and generates a sequence of functions \begin{align*}\left \{Y_n(x) \right \}\end{align*} which converges to the initial value problem. To summarize this procedure mathematically,

Picard’s Method

Let \begin{align*}\{Y_{n}(x)\}\end{align*} be sequence defined successively by,

\begin{align*}Y_n(x) = y_0 + \int_{x_0}^xf(t, Y_{n - 1} (t)) dt\end{align*} for \begin{align*}n \ge 0\end{align*}

The sequence of approximations converges to the solution \begin{align*}y(x)\end{align*}, i.e.

\begin{align*}\lim_{n \to \infty} Y_n (x) = y(x).\end{align*}

Now that we have defined Picard’s method, let’s calculate a sequence of functions for an initial value problem.

Example 1

Find the first four functions \begin{align*}\left \{Y_n (x)\right \}^3_{n = 0}\end{align*} defined by Picard’s method for the solution to the initial value problem

\begin{align*}y'(x) = xy(x\end{align*}) with \begin{align*}y(-1) = 1\end{align*}.


We want to apply the Fundamental Theorem of Calculus to the differential equations so that it is reformulated for use in the Picard method. Thus,

\begin{align*}\int^x_{-1} y'(t) dt & = \int^x_{-1} ty(t)dt\\ y(x) - y(-1) & = \int^x_{-1} ty(t)dt\\ y(x) & = 1 + \int^x_{-1} ty(t)dt\end{align*}

Now that the differential equation has been rewritten for Picard’s method, we begin the calculations for the sequence of functions. In all cases the first function \begin{align*}Y_0(x)\end{align*} is given by the initial condition:

Step 1 - Define \begin{align*}Y_0(x) = 1\end{align*}

Step 2 - Substitute \begin{align*}Y_0(x) = 1\end{align*} for \begin{align*}y(t)\end{align*} in the integrand of \begin{align*}y(x) = 1 + \int^x_{-1} ty(t)dt\end{align*}:

\begin{align*}Y_1(x) & = 1 + \int^x_{-1} t dt \\ Y_1(x) & = 1 + \frac{t^2} {2} \Bigg|^x_{-1} \\ Y_1(x) & = \frac{1} {2} + \frac{x^2} {2}\end{align*}

Step 3 - Substitute \begin{align*}Y_1(x) = \frac{1} {2} + \frac{x^2} {2}\end{align*} for \begin{align*}y(t)\end{align*} in the integrand as above:

\begin{align*}Y_2(x) & = 1 + \int^x_{-1}t\left (\frac{1} {2} + \frac{t^2} {2}\right )dt \\ Y_2(x) & = 1 + \left (\frac{t^2} {4} + \frac{t^4} {8}\right ) \Bigg|^x_{-1} \\ Y_2(x) & = \frac{5} {8} + \frac{x^2} {4} + \frac{x^4} {8}\end{align*}

Step 4 - Substitute \begin{align*}Y_2(x) = \frac{5} {8} + \frac{x^2} {4} + \frac{x^4} {8}\end{align*} for \begin{align*}y(t)\end{align*} in the integrand as done previously:

\begin{align*}Y_3 (x) & = 1 + \int^x_{-1}t \left(\frac{5} {8} + \frac{t^2} {4} + \frac{t^4} {8}\right )dt \\ Y_3 (x) & = 1 + \left(\frac{5t^2} {16} + \frac{t^4} {16} + \frac{t^6} {48}\right ) \Bigg|^x_{-1} \\ Y_3 (x) & = \frac{29} {48} + \frac{5x^2} {16} + \frac{x^4} {16} + \frac{x^6} {48}\end{align*}

Thus, the initial four functions in the sequence defined by Picard’s method are:

\begin{align*}\left \{1, \frac{1} {2} + \frac{x^2} {2}, \quad \quad \frac{5} {8} + \frac{x^2} {4} + \frac{x^4} {8}, \quad \quad \frac{29} {48} + \frac{5x^2} {16} + \frac{x^4} {16} + \frac{x^6} {48}\right \}\end{align*}

The method also states that this sequence will converge to the solution \begin{align*}y(x)\end{align*} of the initial value problem, i.e.

\begin{align*}\lim_{n \to +\infty} Y_n(x) = y(x)\end{align*}

A pattern of the functions in the sequence \begin{align*}Y_n(x)\end{align*} is emerging but it is not an obvious one. We do know \begin{align*}Y_n(x)\end{align*} will converge to the solution for this problem by Picard’s method. The exact solution for this problem can be calculated and is given by:

\begin{align*}y(x) = e^{\frac{x^2 - 1} {2}}\end{align*}

Clearly this solution satisfies \begin{align*}y(x) = xy(x)\end{align*} and \begin{align*}y(-1) = 1\end{align*}.

Review Questions

  1. Find the rule for the sequence \begin{align*}a_n\end{align*}. \begin{align*}& n & & 1 & & 2 & & 3 & & 4 & & \ldots\\ & a_n = ? & & -2 & & 2 & & -2 & & 2 & & \ldots\end{align*}

Tell if each sequence is convergent, is divergent, or has no limit. If the sequence is convergent, find its limit.

  1. \begin{align*}\left \{\frac{4} {n} + \frac{3} {n^2}\right \}\end{align*}
  2. \begin{align*}\left \{6 - \frac{7} {\sqrt{n}}\right \}\end{align*}
  3. \begin{align*}-5, 5, -5, 5, -5, 5, \ldots\end{align*}
  4. \begin{align*}\left \{\frac{4n^6 - 7} {3n}\right \}\end{align*}
  5. \begin{align*}\left \{\frac{(-1)^n} {5n^2}\right \}\end{align*}
  6. \begin{align*}\left \{(-1)^n n \right \}\end{align*}
  7. \begin{align*}\left \{(-1)^n \frac{3n^4 - 2} {2n^4 + 6n^2 - 4n}\right \}\end{align*}
  8. \begin{align*}\left \{\frac{6n^2} {e^n}\right \}\end{align*}
  9. Let \begin{align*}\left \{a_n \right \}\end{align*} be a sequence such that \begin{align*}\lim_{n \to +\infty} |a_n| = 0\end{align*}. Show that \begin{align*}\lim_{n \to +\infty} a_n = 0\end{align*}. (\begin{align*}| a_n |\end{align*} is the absolute value of \begin{align*}a_n\end{align*}.)
  10. Find the first four functions \begin{align*}\left \{ Y_n (x)\right \}^3_{n = 0}\end{align*} defined by Picard’s method for the solution to the initial value problem \begin{align*}y'(x) = 1 + y\end{align*} with \begin{align*}y(0) = 0\end{align*}.
  11. Find the first four functions \begin{align*}\left \{ Y_n (x)\right \}^3_{n = 0}\end{align*} defined by Picard’s method for the solution to the initial value problem \begin{align*}y'(x) = 1 + y^2\end{align*} with \begin{align*}y(0) = 0\end{align*}.
  12. Find the first three functions \begin{align*}\left \{ Y_n (x)\right \}^2_{n = 0}\end{align*} defined by Picard’s method for the solution to the initial value problem \begin{align*}y'(x) = y^{1/3}\end{align*} with \begin{align*}y(0) = \frac{1} {8}\end{align*}.

Review Answers

  1. \begin{align*}a_n = (-1)^n2\end{align*}
  2. convergent; Limit is
  3. convergent; Limit is \begin{align*}6\end{align*}
  4. No limit exists.
  5. divergent
  6. convergent; Limit is
  7. No limit exists.
  8. No limit exists.
  9. convergent; Limit is
  10. By definition of absolute value, \begin{align*}-| a_n | \le a_n \le | a_n |\end{align*} . Then take limits of all three terms: \begin{align*}\lim_{n \to +\infty} (-1|a_n|) & \le \lim_{n \to +\infty} a_n \le \lim_{n \to +\infty} |a_n|\\ -\lim_{n \to +\infty} (|a_n|) & \le \lim_{n \to +\infty} a_n \le \lim_{n \to +\infty} |a_n|\\ 0 & \le \lim_{n \to +\infty} a_n \le 0\end{align*}

By the Sandwich/Squeeze Theorem, \begin{align*}\lim_{n \to +\infty} a_n = 0\end{align*} also.

  1. \begin{align*}\left \{Y_n (x)\right \}^3_{n = 0} = \left \{0, x, x + \frac{x^2} {2}, x + \frac{x^2} {2} + \frac{x^3} {6}\right \}\end{align*}
  2. \begin{align*}\left \{Y_n (x)\right \}^3_{n = 0} = \left \{0, x, x + \frac{x^3} {3}, x + \frac{x^3} {3} + \frac{2x^5} {15} + \frac{x^7} {63}\right \}\end{align*}
  3. \begin{align*}\left \{Y_n (x)\right \}^2_{n = 0} = \left \{\frac{1} {8}, \frac{1} {8} + \frac{x} {2}, \frac{1} {32} + \frac{3} {2} \left (\frac{1} {8} + \frac{x} {2}\right )^{4/3} \right \}\end{align*}


  1. sequence
  2. rules
  3. terms
  4. index, indices
  5. limit
  6. convergence
  7. divergence
  8. L’Hôpital’s Rule
  9. Sandwich/Squeeze Theorem
  10. Picard’s Method

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