10.2: Infinite Series and Divergence Test

In Section 10.1 we discussed infinite sequences. We now investigate infinite series. Series are fundamental to model the world when ordinary functions fall short, and we investigate various types of series and their convergence or divergence. We cover the following topics:

• Defining an Infinite Series
• Divergence Test
• Infinite Geometric Series
• Telescoping Series
• Properties of Infinite Series

Defining an Infinite Series

A series is the sum of a pattern of numbers. For example, consider the sequence \[\{a_n\} = \{1, 2, 3, 4, 5\} \pd\] Then a series with \(a_n\) is \[1 + 2 + 3 + 4 + 5 = \sum_{n = 1}^5 a_n \pd\] An infinite series is the sum of infinitely many numbers—namely, \begin{equation} S = a_1 + a_2 + a_3 + \cdots = \sum_{n = 1}^\infty a_n \pd \label{eq:inf-series} \end{equation} It may seem paradoxical to add infinitely many numbers; shouldn't the sum be \(\infty\) or \(-\infty \ques\) The answer is not always; we'll see that if we add numbers whose magnitudes decrease to \(0,\) then the sum may approach a finite value.

A partial sum approximates an infinite series to a finite number of terms; examples of partial sums to \(\eqref{eq:inf-series}\) are \[ \ba S_1 &= a_1 \nl S_2 &= a_1 + a_2 \nl S_3 &= a_1 + a_2 + a_3 \pd \ea \] The general, \(N\)th partial sum to \(\eqref{eq:inf-series}\) is \begin{equation} S_N = a_1 + a_2 + a_3 + \cdots + a_N = \sum_{i = 1}^N a_i \pd \label{eq:partial-sum} \end{equation} As we add more terms to a partial sum, it becomes closer to the sum of infinitely many terms, assuming the infinite sum exists. In \(\eqref{eq:partial-sum}\) allowing \(N\) to grow arbitrarily enables \(\sum_{i = 1}^N a_i\) to become a better and better approximation to \(\sum_{n = 1}^\infty a_n.\) Thus, \(\sum_{n = 1}^\infty a_n\) is the limiting value of \(\sum_{i = 1}^N a_i\) as \(N \to \infty;\) that is, \[\sum_{n = 1}^\infty a_n = \lim_{N \to \infty} \sum_{i = 1}^N a_i \pd\]

If \(\lim_{N \to \infty} \sum_{i = 1}^N a_i\) exists, then \(\sum_{n = 1}^\infty a_n\) is called convergent. But if the limit of the partial sum does not exist, then \(\sum_{n = 1}^\infty a_n\) is called divergent. In words, an infinite series is convergent if adding more terms allows the sum to approach a fixed number, and divergent if the sum does not approach a final number as we add more terms.

Divergence Test

Consider the series \[S = 1 + 2 + 3 + \cdots \pd\] As more terms are added, \(S\) grows arbitrarily and does not settle to a fixed number; hence, we say the series is divergent (or that the series diverges). The same conclusion applies to \[S = 1 + 1 + 1 + \cdots \pd\] We infer the following rule: If the terms of an infinite series do not approach \(0,\) then the infinite series must be divergent. Mathematically, we say \(\sum_{n = 1}^\infty a_n\) diverges if \(\lim_{n \to \infty} a_n \ne 0\) or if the limit does not exist. This rule is called the Divergence Test (or \(\bf n\)th Term Test).

We can also express the Divergence Test as follows: \[\lim_{n \to \infty} a_n = 0 \if \sum_{n = 1}^\infty a_n \; \textrm{converges} \pd\] Accordingly, if the series \(\sum_{n = 1}^\infty a_n\) converges, then the corresponding sequence \(\{a_n\}\) converges to \(0.\) (The converse of this statement is not necessarily true, so be careful with this test.)

PROOF Let \(S = \sum_{n = 1}^\infty a_n.\) Consider the \(N\)th partial sum of \(S \col\) \[S_N = \sum_{i = 1}^N a_i \pd\] If \(\sum_{n = 1}^\infty a_n\) is convergent, then as \(N\) grows to infinity, \(S_N\) approaches some value and so the difference \(S_N - S_{N - 1}\) becomes \(0.\) We see \[ \ba S_N - S_{N - 1} &= \sum_{i = 1}^N a_i - \sum_{i = 1}^{N - 1} a_i \nl &= (a_1 + a_2 + a_3 + \cdots + a_N) - (a_1 + a_2 + a_3 + \cdots + a_{N - 1}) \nl &= a_N \pd \ea \] But if \(\lim_{n \to \infty} a_n \ne 0\) or the limit does not exist, then the series cannot converge; hence, it diverges. \[\qedproof\]

If \(\lim_{n \to \infty} a_n = 0,\) then the Divergence Test gives no information about whether \(\sum_{n = 1}^\infty a_n\) converges or diverges. In other words, the condition \(\lim_{n \to \infty} a_n = 0\) is a necessary but insufficient condition for convergence. For example, the series \[1 + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = \sum_{n = 1}^\infty \frac{1}{n^2}\] turns out to be convergent. But the series \begin{equation} 1 + \frac{1}{2} + \frac{1}{3} + \cdots = \sum_{n = 1}^\infty \frac{1}{n} \cma \label{eq:harmonic-series} \end{equation} called the Harmonic series, is divergent. (We will prove this fact in Section 10.3.) Both \(1/n^2\) and \(1/n\) approach \(0\) as \(n \to \infty,\) so the Divergence Test is inconclusive about whether each series converges or diverges.

Infinite Geometric Series

It is rare to be able to find the exact sum of a convergent series. In this chapter, we primarily investigate methods to determine whether an infinite series has a finite sum. We discuss two types of infinite series whose exact sums we can calculate, however: geometric series and telescoping series.

A geometric series is a sum in which each term is obtained by multiplying the previous term by a constant number. We define a finite geometric sum of \(N\) terms as \[S_N = a + ar + ar^2 + \cdots + ar^{N - 1} = \sum_{i = 0}^{N - 1} ar^i \cma\] where \(r\) is called the common ratio. To find its sum, we multiply the terms of \(S_N\) by \(r\) and subtract the result from the terms of \(S_N.\) Assuming \(r \ne 1,\) we have \begin{align} S_N &= a + \cancel{ar} + \cancel{ar^2} + \cdots + \cancel{ar^{N - 1}} \nonum \nl -(rS_N &= \cancel{ar} + \cancel{ar^2} + \cancel{ar^3} + \cdots + ar^{N}) \nonum \nl S_N - rS_N &= a - ar^N \nonumber \nl S_N(1 - r) &= a(1 - r^N) \nonumber \nl \implies S_N &= \frac{a(1 - r^N)}{1 - r} \pd \label{eq:finite-geo-series} \end{align} We define an infinite geometric series as follows: \begin{equation} S = \sum_{n=0}^\infty ar^n = a + ar + ar^2 + \cdots \pd \label{eq:geo-series} \end{equation} To find the sum \(S,\) we allow \(N\) to grow arbitrarily large in \(\eqref{eq:finite-geo-series}.\) Thus, we represent \(S\) as \(\lim_{N \to \infty} S_N\) if the limit exists. In \(\eqref{eq:finite-geo-series}\) if \(\abs r \lt 1,\) then \(r^N \to 0\) as \(N \to \infty.\) So \[\lim_{N \to \infty} S_N = \frac{a}{1 - r} \cma\] meaning \(S\) is finite and so the geometric series is convergent. By contrast, if \(\abs r \geq 1,\) then \(\lim_{N \to \infty} S_N\) doesn't exist, so \(\sum_{n = 0}^\infty ar^n\) diverges by the Divergence Test. Accordingly, an infinite geometric series converges for \(r\) in the interval \((-1, 1)\) (called the interval of convergence), and its sum is given by \begin{equation} \sum_{n = 0}^\infty ar^n = \frac{a}{1 - r} \cmaa \abs r \lt 1 \pd \label{eq:inf-geo-series} \end{equation} It is easiest to think of the result of \(\eqrefer{eq:inf-geo-series}\) as the first term divided by \((1 - r).\) This result is intuitive: If \(\abs r \lt 1,\) then the terms are decreasing in magnitude to \(0\) and so the sum \(S\) is finite. But if \(\abs r \geq 1,\) then the terms do not approach \(0,\) so the series \(\sum_{n = 0}^\infty ar^n\) is divergent.

Telescoping Series

In a telescoping series, many of the terms cancel out. As an example, consider the following partial sum: \begin{equation} \ba S_N &= \sum_{i = 1}^N \par{a_i - a_{i + 1}} \nl &= \par{a_1 - a_2} + \par{a_2 - a_3} + \cdots + \par{a_{N - 2} - a_{N - 1}} + \par{a_{N - 1} - a_N} + \par{a_N - a_{N + 1}} \nl &= a_1 - a_{N + 1} \pd \ea \label{eq:telescoping-sum} \end{equation} In this form, the infinite sum \(\sum_{n = 1}^\infty \par{a_n - a_{n + 1}}\) (if it converges) is easily found by taking \(\lim_{N \to \infty} S_N.\)

When the summation function \(a_i\) is a rational expression, it is difficult to inspect whether \(\sum a_i\) is a telescoping series. Therefore, partial fraction decomposition (see Section 6.4) may be needed to reexpress \(a_i\) as a difference of fractions. Most telescoping series are not obvious, so it's helpful to write out terms and try to find a pattern of canceling.

Properties of Infinite Series

We now discuss the addition and subtraction of infinite series, as well as multiplying a series by a constant. Note the following properties.

PROOF Let the \(N\)th partial sums of \(\sum_{n = 1}^\infty a_n\) and \(\sum_{n = 1}^\infty b_n\) be, respectively, \[\sum_{i = 1}^N a_i \and \sum_{i = 1}^N b_i \pd\] The \(N\)th partial sum of \(\sum_{n = 1}^\infty (a_n + b_n)\) is given by \[\sum_{i = 1}^N (a_i + b_i) \pd\] Taking the limit as \(N\) approaches infinity, we attain \[ \ba \lim_{N \to \infty} \sum_{i = 1}^N \par{a_i + b_i} &= \lim_{N \to \infty} \sum_{i = 1}^N a_i + \lim_{N \to \infty} \sum_{i = 1}^N b_i \nl &= \sum_{n = 1}^\infty a_n + \sum_{n = 1}^\infty b_n \pd \ea \] The last step is true because \(\sum_{n = 1}^\infty a_n\) and \(\sum_{n = 1}^\infty b_n\) are convergent series (see Defining an Infinite Series). Thus, \[\sum_{n = 1}^\infty (a_n + b_n) = \sum_{n = 1}^\infty a_n + \sum_{n = 1}^\infty b_n \pd\] The proof for (b) is nearly identical if we consider \(\sum_{n = 1}^\infty (a_n - b_n).\) \[\qedproof\]

PROOF Let \(S = \sum_{n = 1}^\infty a_n.\) Also let \(T = c + S;\) the \(N\)th partial sum of \(T\) is \[T_N = c + \sum_{i = 1}^N a_i \pd\] Taking the limit of \(T_N\) as \(N \to \infty\) gives \[T = \lim_{N \to \infty} \parbr{c + \sum_{i = 1}^N a_i} \pd\] Using the Sum Law of Limits (see Section 1.2), we attain \[T = \lim_{N \to \infty} c + \lim_{N \to \infty} \sum_{i = 1}^N a_i = c + \lim_{N \to \infty} \sum_{i = 1}^N a_i \pd\] If \(\sum_{n = 1}^\infty a_n\) converges, then \(\sum_{i = 1}^N a_i\) approaches \(S\) as \(N\) grows to infinity, so \(T\) converges to \(c + S.\) But if \(\sum_{n = 1}^\infty a_n\) diverges, then \(\lim_{N \to \infty} \sum_{i = 1}^N a_i\) doesn't exist. Accordingly, \(T\) does not approach a finite value as \(N \to \infty,\) so \(c + \sum_{n = 1}^\infty a_n\) diverges. This same process applies to prove the convergence or divergence of \(T = c - S.\) Proving part (c) is left as an exercise. \[\qedproof\]

Because of parts (b) and (c), the starting index of an infinite series does not change its convergence or divergence. For example, if \(a_n\) is defined for \(n \geq 1\) and \(\sum a_n\) converges, then \[\andThree{\sum_{n = 1}^\infty a_n}{\sum_{n = 3}^\infty a_n}{\sum_{n = 100}^\infty a_n}\] are all convergent because they only vary by a finite number of terms (whose sums are finite). Adding a constant to a finite sum yields a finite result, and multiplying a finite sum by a constant produces a finite answer. As a result, the following example series all converge: \[\andThree{2 + \sum_{n = 1}^\infty a_n}{-6 + \sum_{n = 1}^\infty a_n}{100 \sum_{n = 1}^\infty a_n} \pd\]