What is the difference between a convergent and divergent series?

A series is convergent if the limit of its partial sums as n approaches infinity exists and is a finite real number. If the limit does not exist or is infinite, the series is divergent.

Convergent and Divergent Series – Calculus Tutorial & Examples

Convergence and divergence of Series are presented with examples and their detailed solutions.

In what follows, \( C \) representsة the constant of integration where applicable.

Definition of Series

Let \( \displaystyle \left\{a_n \right\}_1^{\infty} \) be an infinite sequence. The infinite sum: \[ a_1 + a_2 + a_3 + \dots + a_n + \dots \] is called a series [1] [2].

The series may also be written using the summation symbol \( \displaystyle \sum \) as: \[ \sum_{i=1}^{\infty} a_i \]

Note that when the upper limit of the sum is infinite as in \( \displaystyle \sum_{i=1}^{\infty} a_i \), we have a series, and when the upper limit of the sum is finite as in \( \displaystyle \sum_{i=1}^{n} a_i \), we have a partial sum.

Convergent and Divergent Series

We first discuss the convergence and divergence of series using graphs and two examples.

Example 1 - Convergent Series

Let \( a_i = \dfrac{1}{4^i} \) and define the partial sums as: \[ S_n = \sum_{i=1}^{n} \dfrac{1}{4^i} \]

The table of values of \( n \), \( a_n \), and the finite sums \( s_n \) defined by:

\( s_1 = a_1 \)
\( s_2 = a_1 + a_2 \)
\( s_3 = a_1 + a_2 + a_3 \)
\( s_4 = a_1 + a_2 + a_3 + a_4 \)

are shown below. Note that as we add more terms to the partial sum \( s_n \), it becomes closer to \( 2 \), as clearly shown in Table 1 and Graph 1 below.

Table of a Convergent Series — Table 1. Table of Values of \( n \), \( a_n \), and the Partial Sums \( \displaystyle \sum_{i=1}^{n} \dfrac{1}{4^i} \)

Graph of a Convergent Series — Graph 1. Graph of Partial Sums \( \displaystyle \sum_{i=1}^{n} \dfrac{1}{4^i} \)

As \( n \) increases, the partial sum \( \displaystyle \sum_{i=1}^{n} \dfrac{1}{4^i} \) tends to a constant value equal to \( 2 \). We can therefore say that the series \( \displaystyle \sum_{i=1}^{\infty} \dfrac{1}{4^i} \) is convergent to \( 2 \).

Example 2 - Divergent Series

Let \( a_i = (-2)^i \) and define the partial sums as: \[ S_n = \sum_{i=1}^{n} (-2)^i \]

Below is the table of values of \( n \), \( a_n \), and the finite sums \( s_n \). As we add more terms, neither Table 2 nor Graph 2 shows convergence to any value. The partial sum has no limit, and we consider the series \( \displaystyle \sum_{i=1}^{\infty} (-2)^i \) to be divergent.

Table of a Divergent Series — Table 2. Table of Values of Partial Sums \( \displaystyle \sum_{i=1}^{n} (-2)^i \)

Graph of a Divergent Series — Graph 2. Graph of Partial Sums \( \displaystyle \sum (-2)^i \)

Formal Definition of Convergent and Divergent Series

Given a series: \[ \sum_{i=1}^{\infty} a_i = a_1 + a_2 + a_3 + \dots \] Let \( S_n \) be the partial sum: \[ S_n = \sum_{i=1}^{n} a_i = a_1 + a_2 + \dots + a_n \]

If \( \lim_{n\to\infty} S_n \) exists and: \[ \lim_{n\to\infty} S_n = s \] where \( s \) is a real number, we say that the series \( \displaystyle \sum_{i=1}^{\infty} a_i \) is convergent and write: \[ \sum_{i=1}^{\infty} a_i = a_1 + a_2 + a_3 + \dots = s \]

If \( \lim_{n\to\infty} S_n \) does not exist or is not a real number, the series \( \sum_{i=1}^{\infty} a_i \) is divergent. We can also write: \[ \sum_{i=1}^{\infty} a_i = \lim_{n\to\infty} \sum_{i=1}^{n} a_i \]

Geometric Series

A geometric series is defined as: \[ \sum_{i=1}^{\infty} a \, r^{i-1} = a + a \, r + a \, r^2 + \dots \] where \( r \) is called the common ratio.

Let \( S_n \) be the partial sum defined by: \[ S_n = a + a \, r + a \, r^2 + \dots + a \, r^{n-1} \quad (I) \] Multiply both sides by \( r \): \[ r \, S_n = a \, r + a \, r^2 + a \, r^3 + \dots + a \, r^n \quad (II) \] Subtract (II) from (I): \[ S_n - r \, S_n = a(1 - r^n) \implies S_n(1 - r) = a(1 - r^n) \implies S_n = \frac{a(1 - r^n)}{1 - r} \]

It is known from limits that \( \lim_{n\to\infty} r^n = 0 \) if \( |r| < 1 \). Thus, we state:

A geometric series \( \displaystyle \sum_{i=1}^{\infty} a r^{i-1} = a + a r + a r^2 + \dots \) is convergent if \( |r| < 1 \), and its sum is given by: \[ \sum_{i=1}^{\infty} a r^{i-1} = \frac{a}{1 - r}, \quad |r| < 1 \] If \( |r| > 1 \), the geometric series is divergent.

Example 3 - Geometric Series

Which of the following geometric series are convergent? Find the sum if possible.

\( 4 - \dfrac{12}{5} + \dfrac{36}{25} - \dfrac{108}{125} + \dots \)
\( \displaystyle \sum_{i=1}^{\infty} \dfrac{3^{2i}}{10^{i-1}} \)
\( \displaystyle \sum_{i=1}^{\infty} \dfrac{4}{5} \cdot \dfrac{1}{5^{-i} \; 2^{i+1}} \)

Solutions to Example 3

The series \( 4 - \dfrac{12}{5} + \dfrac{36}{25} - \dfrac{108}{125} + \dots \) has common ratio \( r = \dfrac{-\frac{12}{5}}{4} = -\dfrac{3}{5} \).
Since \( |r| = \dfrac{3}{5} < 1 \), the series is convergent.
With first term \( a = 4 \), the sum is: \[ \sum = \frac{4}{1 - (-\frac{3}{5})} = \frac{4}{1 + \frac{3}{5}} = \frac{4}{\frac{8}{5}} = \frac{5}{2} \]
Rewrite the series: \[ \sum_{i=1}^{\infty} \frac{3^{2i}}{10^{i-1}} = \sum_{i=1}^{\infty} \frac{(3^2)^i}{10^{i-1}} = \sum_{i=1}^{\infty} \frac{9 \cdot 9^{i-1}}{10^{i-1}} = \sum_{i=1}^{\infty} 9 \left(\frac{9}{10}\right)^{i-1} \] First term \( a = 9 \), common ratio \( r = \dfrac{9}{10} < 1 \). The series is convergent: \[ \sum = \frac{9}{1 - \frac{9}{10}} = 90 \]
Rewrite the series: \[ \sum_{i=1}^{\infty} \frac{4}{5} \cdot \frac{1}{5^{-i} \, 2^{i+1}} = \sum_{i=1}^{\infty} \frac{2^2}{5^1} \cdot \frac{5^i}{2^{i+1}} = \sum_{i=1}^{\infty} \frac{5^{i-1}}{2^{i-1}} = \sum_{i=1}^{\infty} \left(\frac{5}{2}\right)^{i-1} \] The common ratio is \( r = \dfrac{5}{2} > 1 \), hence the series is divergent.

Arithmetic Series Are Divergent

An arithmetic series is given by: \[ \sum_{i=1}^{\infty} (a + (i-1)d) = a + (a + d) + (a + 2d) + \dots \] The partial sum \( S_n \) is: \[ S_n = \frac{n(2a + (n-1)d)}{2} = \frac{nd}{2}n + n\left(a - \frac{d}{2}\right) \] Since \( \lim_{n\to\infty} S_n = \infty \), all arithmetic series are divergent.

Theorem of Combinations of Convergent Series

If \( \displaystyle \sum_{n=1}^{\infty} a_n = L_1 \) and \( \displaystyle \sum_{n=1}^{\infty} b_n = L_2 \) are convergent series, and \( k_1, k_2 \) are real numbers, then any linear combination is convergent: \[ k_1 \sum_{n=1}^{\infty} a_n + k_2 \sum_{n=1}^{\infty} b_n = k_1 L_1 + k_2 L_2 \]

Example 4

Evaluate, if possible, the following series:

\( \left(2 - 1 + \dfrac{1}{2} - \dfrac{1}{4} + \dots\right) + \left(-\dfrac{1}{3} + \dfrac{1}{9} - \dfrac{1}{27} + \dots\right) \)
\( -3 \sum_{i=1}^{\infty} \left(\dfrac{1}{2}\right)^{i-1} - 4 \sum_{i=1}^{\infty} 0.7^{i-1} \)
\( \sum_{i=1}^{\infty} (0.2)^i + \sum_{i=1}^{\infty} \dfrac{3^i - 7 \cdot 2^i}{5^{i-1}} \)

Solutions to Example 4

First series: \( A_1 = 2, R_1 = -\dfrac{1}{2} \implies \dfrac{2}{1 - (-1/2)} = \dfrac{4}{3} \).
Second series: \( A_2 = -\dfrac{1}{3}, R_2 = -\dfrac{1}{3} \implies \dfrac{-1/3}{1 - (-1/3)} = -\dfrac{1}{4} \).
Sum = \( \dfrac{4}{3} - \dfrac{1}{4} = \dfrac{13}{12} \).
First series: \( A_1 = 1, R_1 = \dfrac{1}{2} \implies \dfrac{1}{1 - 1/2} = 2 \).
Second series: \( A_2 = 1, R_2 = 0.7 \implies \dfrac{1}{1 - 0.7} = \dfrac{10}{3} \).
Total sum = \( -3(2) - 4\left(\dfrac{10}{3}\right) = -6 - \dfrac{40}{3} = -\dfrac{58}{3} \).
Rewrite as a combination of geometric series: \[ 0.2 \sum_{i=1}^{\infty} (0.2)^{i-1} + 3 \sum_{i=1}^{\infty} \left(\frac{3}{5}\right)^{i-1} - 14 \sum_{i=1}^{\infty} \left(\frac{2}{5}\right)^{i-1} \] Evaluating each sum: \[ 0.2\left(\frac{1}{1-0.2}\right) + 3\left(\frac{1}{1-3/5}\right) - 14\left(\frac{1}{1-2/5}\right) = \frac{1}{4} + \frac{15}{2} - \frac{70}{3} = -\frac{187}{12} \]