MechForge · Calculus · Chapter 11: Infinite Sequences and Series

SECTION 11.1Sequences

A sequence is an ordered, never-ending list of numbers. Because every positive integer is paired with exactly one term, a sequence is really a function whose domain is the positive integers — we simply write the output as a subscripted term rather than in function notation.

Sequence

An infinite sequence is a function defined on the positive integers, listed in order as its terms. The value paired with the integer \(n\) is the \(n\)th term \(a_n\), and the whole list is written \(\{a_n\}\) or \(\{a_n\}_{n=1}^{\infty}\). Each term \(a_n\) has a well-defined successor \(a_{n+1}\), so the list goes on without end.

Limit of a Sequence (informal)

A sequence \(\{a_n\}\) has limit \(L\), written \(\lim_{n\to\infty}a_n=L\), if its terms can be made as close to \(L\) as we please by taking \(n\) large enough. A sequence with a finite limit is convergent; otherwise it is divergent.

Precise definition of a sequence limit

\[ \lim_{n\to\infty}a_n=L \iff \text{for every }\varepsilon>0\text{ there is an integer }N\text{ with }|a_n-L|<\varepsilon\text{ whenever }n>N. \]

Figure 11.1.1 The terms of a convergent sequence eventually enter and stay inside every horizontal band of half-width \(\varepsilon\) around the limit \(L\).

Limit laws and key theorems

The familiar limit laws for functions carry over to sequences: limits of sums, differences, products, quotients, and powers behave exactly as expected, provided the individual limits exist (and denominators are nonzero in the limit).

Function-to-sequence limit

If \(\lim_{x\to\infty}f(x)=L\) and \(f(n)=a_n\) for every integer \(n\), then \(\lim_{n\to\infty}a_n=L\). This lets us bring tools such as l'Hospital's Rule — which apply to functions of a real variable — to bear on sequences.

Squeeze Theorem for sequences

If \(a_n\le b_n\le c_n\) for all \(n\ge n_0\) and \(\lim_{n\to\infty}a_n=\lim_{n\to\infty}c_n=L\), then \(\lim_{n\to\infty}b_n=L\). A useful companion: if \(\lim_{n\to\infty}|a_n|=0\), then \(\lim_{n\to\infty}a_n=0\).

Continuous-function theorem

If \(\lim_{n\to\infty}a_n=L\) and \(f\) is continuous at \(L\), then \(\lim_{n\to\infty}f(a_n)=f(L)\). Applying a continuous function to a convergent sequence preserves convergence.

Two standard sequence limits

\[ \lim_{n\to\infty}\frac{1}{n^{r}}=0\ \ (r>0); \qquad \lim_{n\to\infty}r^{n}=\begin{cases}0 & -1

EXAMPLE 1 A rational-term limit

Evaluate \(\displaystyle\lim_{n\to\infty}\frac{n}{n+1}\).

Solution

Divide numerator and denominator by the highest power of \(n\): \(\dfrac{n}{n+1}=\dfrac{1}{1+\tfrac1n}\).

As \(n\to\infty\), \(\tfrac1n\to0\), so the expression tends to \(\dfrac{1}{1+0}=1\). The sequence converges to \(1\).

EXAMPLE 2 Using l'Hospital via a related function

Find \(\displaystyle\lim_{n\to\infty}\frac{\ln n}{n}\).

Solution

The discrete ratio is hard to attack directly, so study the companion function \(f(x)=\dfrac{\ln x}{x}\).

By l'Hospital's Rule, \(\displaystyle\lim_{x\to\infty}\frac{\ln x}{x}=\lim_{x\to\infty}\frac{1/x}{1}=0.\)

Since \(f(n)=a_n\), the function-to-sequence theorem gives \(\displaystyle\lim_{n\to\infty}\frac{\ln n}{n}=0.\)

EXAMPLE 3 A divergent oscillating sequence

Is \(a_n=(-1)^{n}\) convergent?

Solution

Written out, the terms are \(-1,1,-1,1,\dots\), bouncing between two values forever.

They never settle near a single number, so the limit does not exist — the sequence diverges.

Monotonic and bounded sequences

Monotonic and bounded

A sequence is increasing if \(a_ndecreasing if \(a_n>a_{n+1}\) for all \(n\); either way it is monotonic. It is bounded above if some \(M\) satisfies \(a_n\le M\) for all \(n\), bounded below if some \(m\) satisfies \(a_n\ge m\), and simply bounded if both hold.

Monotonic Sequence Theorem

Every bounded, monotonic sequence is convergent. The theorem guarantees a limit exists without telling us its value — but once existence is known, a recurrence relation can often be solved to find it.

The Monotonic Sequence Theorem is the one existence result you'll lean on again and again when working with recursively defined sequences. The two-step move is: (1) prove the sequence is monotonic and bounded by induction, then (2) take the limit of both sides of the recurrence \(a_{n+1}=g(a_n)\) and solve \(L=g(L)\). Practice this pairing until it's automatic.

EXAMPLE 4 Limit of a recursive sequence

A sequence is defined by \(a_1=2\) and \(a_{n+1}=\tfrac12(a_n+6)\). Investigate its convergence.

Solution

Computing terms gives \(2,4,5,5.5,5.75,\dots\) — the values rise toward \(6\), suggesting the sequence is increasing and bounded above by \(6\).

Induction confirms both: if \(a_k

By the Monotonic Sequence Theorem the limit \(L\) exists. Taking limits of the recurrence: \(L=\tfrac12(L+6)\), hence \(L=6\).

SECTION 11.2Series

Adding the terms of an infinite sequence produces an infinite series. We give meaning to such an "infinite sum" not by adding term by term forever, but by tracking the sequence of partial sums and asking whether it converges.

Infinite series and partial sums

Given a sequence \(\{a_n\}\), the expression \(\sum_{n=1}^{\infty}a_n\) is an infinite series. Its \(n\)th partial sum is \(s_n=a_1+a_2+\cdots+a_n=\sum_{i=1}^{n}a_i\). The series converges with sum \(s\) if the sequence \(\{s_n\}\) converges to the finite number \(s\); otherwise the series diverges.

Sum of a series as a limit of partial sums

\[ \sum_{n=1}^{\infty}a_n=s \iff \lim_{n\to\infty}s_n=s,\qquad s_n=\sum_{i=1}^{n}a_i. \]

Figure 11.2.1 A geometric picture of \(\tfrac12+\tfrac14+\tfrac18+\cdots=1\): each piece halves the remaining area, so the partial sums creep up to fill the whole unit.

The geometric series

Geometric series

The series \(\sum_{n=1}^{\infty}ar^{\,n-1}=a+ar+ar^2+\cdots\) (with \(a\ne0\)) converges when \(|r|<1\) and diverges when \(|r|\ge1\). When it converges its sum is the first term divided by one minus the common ratio.

Sum of a convergent geometric series

\[ \sum_{n=1}^{\infty}ar^{\,n-1}=\frac{a}{1-r},\qquad |r|<1. \]

EXAMPLE 1 Summing a geometric series

Find the sum of \(5-\tfrac{10}{3}+\tfrac{20}{9}-\tfrac{40}{27}+\cdots\).

Solution

The first term is \(a=5\) and the common ratio is \(r=-\tfrac23\), so \(|r|<1\) and the series converges.

Apply the formula: \(\dfrac{a}{1-r}=\dfrac{5}{1-(-\tfrac23)}=\dfrac{5}{\tfrac53}=3.\)

EXAMPLE 2 A repeating decimal as a fraction

Write \(2.3\overline{17}=2.3171717\ldots\) as a ratio of integers.

Solution

Separate the non-repeating part: \(2.3+\dfrac{17}{10^{3}}+\dfrac{17}{10^{5}}+\cdots\), where the tail is geometric with \(a=\tfrac{17}{1000}\) and \(r=\tfrac{1}{100}\).

Sum the tail: \(\dfrac{17/1000}{1-1/100}=\dfrac{17}{990}\).

Add the integer-fraction part: \(2.3+\dfrac{17}{990}=\dfrac{23}{10}+\dfrac{17}{990}=\dfrac{1147}{495}.\)

Telescoping sums and the harmonic series

EXAMPLE 3 A telescoping series

Show that \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\) converges and find its sum.

Solution

Use partial fractions: \(\dfrac{1}{n(n+1)}=\dfrac1n-\dfrac1{n+1}\).

The partial sum collapses: \(s_n=\left(1-\tfrac12\right)+\left(\tfrac12-\tfrac13\right)+\cdots+\left(\tfrac1n-\tfrac1{n+1}\right)=1-\dfrac1{n+1}.\)

Hence \(\displaystyle\lim_{n\to\infty}s_n=1\), so the series converges to \(1\).

The harmonic series diverges

The series \(\sum_{n=1}^{\infty}\tfrac1n=1+\tfrac12+\tfrac13+\cdots\) diverges, even though its terms shrink to zero. Grouping terms in blocks whose sums each exceed \(\tfrac12\) shows the partial sums grow without bound. This is the classic warning that small terms are not enough to force convergence.

Test for Divergence

If \(\sum a_n\) converges, then \(\lim_{n\to\infty}a_n=0\). Contrapositively, if \(\lim_{n\to\infty}a_n\) fails to exist or is nonzero, the series diverges. Caution: \(a_n\to0\) alone does not guarantee convergence (the harmonic series is the counterexample).

Algebra of convergent series

If \(\sum a_n\) and \(\sum b_n\) converge and \(c\) is a constant, then \(\sum ca_n\), \(\sum(a_n+b_n)\), and \(\sum(a_n-b_n)\) all converge, with \(\sum ca_n=c\sum a_n\) and \(\sum(a_n\pm b_n)=\sum a_n\pm\sum b_n\). Adding or removing a finite number of terms never changes whether a series converges.

SECTION 11.3The Integral Test and Estimates of Sums

For most series no tidy formula for the partial sum exists, so we test convergence indirectly. The Integral Test compares a series of positive, decreasing terms with the area under a matching continuous curve.

Figure 11.3.1 Rectangles of width 1 sit below a decreasing curve. If the area under the curve is finite, the boxed series is squeezed below it and converges; if infinite, the series diverges with it.

The Integral Test

Let \(f\) be continuous, positive, and decreasing on \([1,\infty)\) with \(a_n=f(n)\). Then \(\sum_{n=1}^{\infty}a_n\) converges if and only if the improper integral \(\int_{1}^{\infty}f(x)\,dx\) converges. The series and the integral share their fate, but their values are generally different.

The \(p\)-Series Test

The \(p\)-series \(\sum_{n=1}^{\infty}\dfrac{1}{n^{p}}\) converges if \(p>1\) and diverges if \(p\le1\). The harmonic series is the borderline case \(p=1\).

EXAMPLE 1 Applying the Integral Test

Test \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}+1}\).

Solution

The function \(f(x)=\dfrac{1}{x^{2}+1}\) is continuous, positive, and decreasing on \([1,\infty)\).

Evaluate the integral: \(\displaystyle\int_{1}^{\infty}\frac{dx}{x^{2}+1}=\lim_{t\to\infty}\big[\tan^{-1}x\big]_1^{t}=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}.\)

The integral is finite, so by the Integral Test the series converges.

EXAMPLE 2 Reading off a p-series

Determine convergence of \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt[3]{n}}\).

Solution

Write \(\dfrac{1}{\sqrt[3]{n}}=\dfrac{1}{n^{1/3}}\), a \(p\)-series with \(p=\tfrac13\).

Since \(p=\tfrac13\le1\), the series diverges.

Remainder estimate for the Integral Test

\[ \int_{n+1}^{\infty}f(x)\,dx \;\le\; R_n \;\le\; \int_{n}^{\infty}f(x)\,dx, \qquad R_n=s-s_n. \]

Don't confuse "the series converges" with "the series equals the integral." The Integral Test only transfers the yes/no verdict. For instance \(\sum 1/n^2=\pi^2/6\) but \(\int_1^\infty x^{-2}\,dx=1\). Use the remainder estimate above when a problem asks how many terms you need for a target accuracy — it's a favourite exam question.

SECTION 11.4The Comparison Tests

When a series of positive terms resembles a known geometric or \(p\)-series, we can settle its convergence by comparing term sizes. Two versions exist: a direct inequality comparison and a more flexible limit comparison.

The Comparison Test

Let \(\sum a_n\) and \(\sum b_n\) be series with positive terms. If \(\sum b_n\) converges and \(a_n\le b_n\) for all \(n\), then \(\sum a_n\) converges. If \(\sum b_n\) diverges and \(a_n\ge b_n\) for all \(n\), then \(\sum a_n\) diverges.

The Limit Comparison Test

Let \(\sum a_n\) and \(\sum b_n\) have positive terms. If \(\displaystyle\lim_{n\to\infty}\frac{a_n}{b_n}=c\) where \(c\) is finite and \(c>0\), then both series converge or both diverge. This avoids the need to engineer a term-by-term inequality.

EXAMPLE 1 Direct comparison

Test \(\displaystyle\sum_{n=1}^{\infty}\frac{5}{2n^{2}+4n+3}\).

Solution

For large \(n\) the denominator behaves like \(2n^2\), so compare with \(\dfrac{5}{2n^2}\).

Since \(2n^2+4n+3>2n^2\), we have \(\dfrac{5}{2n^2+4n+3}<\dfrac{5}{2n^2}\).

The comparison series is a constant times the convergent \(p\)-series (\(p=2\)), so the original series converges.

EXAMPLE 2 Limit comparison with a p-series

Test \(\displaystyle\sum_{n=1}^{\infty}\frac{2n^{2}+3n}{\sqrt{5+n^{5}}}\).

Solution

Keep only the dominant powers: numerator \(\sim 2n^2\), denominator \(\sim n^{5/2}\). This suggests \(b_n=\dfrac{2n^2}{n^{5/2}}=\dfrac{2}{n^{1/2}}\).

The ratio \(\dfrac{a_n}{b_n}\to1>0\) as \(n\to\infty\).

Now \(\sum b_n\) is a constant times a \(p\)-series with \(p=\tfrac12\le1\), which diverges; therefore the original series diverges.

The fast way to pick a comparison series: strip everything except the highest power of \(n\) on top and bottom. The resulting \(b_n\) is almost always a \(p\)-series or geometric series whose behaviour you already know. The Limit Comparison Test then does the rest without fussing over inequalities.

SECTION 11.5Alternating Series

An alternating series has terms that switch sign from one to the next. Such series are far more forgiving than positive series: if the absolute terms simply decrease to zero, the series converges.

Alternating series

An alternating series has the form \(\sum(-1)^{n-1}b_n\) or \(\sum(-1)^{n}b_n\) with \(b_n>0\); the factor \((-1)^{n}\) is what makes consecutive terms alternate in sign.

The Alternating Series Test

If the alternating series \(\sum_{n=1}^{\infty}(-1)^{n-1}b_n\) (with \(b_n>0\)) satisfies (i) \(b_{n+1}\le b_n\) for all \(n\) and (ii) \(\lim_{n\to\infty}b_n=0\), then the series converges.

Figure 11.5.1 Partial sums of a convergent alternating series step back and forth across the limit \(s\), each step shorter than the last because \(b_n\to0\).

EXAMPLE 1 The alternating harmonic series

Does \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=1-\tfrac12+\tfrac13-\tfrac14+\cdots\) converge?

Solution

Here \(b_n=\tfrac1n\). It decreases since \(\tfrac1{n+1}<\tfrac1n\), and \(\lim_{n\to\infty}\tfrac1n=0\).

Both conditions hold, so by the Alternating Series Test the series converges. (Its sum is \(\ln 2\).)

Alternating Series Estimation Theorem

If \(s=\sum(-1)^{n-1}b_n\) satisfies the Alternating Series Test, then the error in using the partial sum \(s_n\) to approximate \(s\) is no larger than the first omitted term: \(|R_n|=|s-s_n|\le b_{n+1}\).

EXAMPLE 2 Approximating a sum to three decimals

Find \(\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}\) correct to three decimal places.

Solution

The terms \(b_n=\tfrac1{n!}\) decrease to zero, so the estimation theorem applies.

Since \(b_7=\tfrac{1}{5040}\approx0.0002\), stopping after the \(n=6\) term guarantees error below \(0.0002\).

Summing the first seven terms gives \(s_6\approx0.368\). (This is precisely \(1/e\).)

SECTION 11.6Absolute Convergence and the Ratio and Root Tests

For series with terms of mixed or irregular sign, the cleanest strategy is to test the series of absolute values. The Ratio and Root Tests then provide automatic machinery, especially for series built from factorials or \(n\)th powers.

Absolute and conditional convergence

A series \(\sum a_n\) is absolutely convergent if \(\sum|a_n|\) converges. It is conditionally convergent if it converges but \(\sum|a_n|\) diverges. The alternating harmonic series is the standard example of conditional convergence.

Absolute convergence implies convergence

If \(\sum|a_n|\) converges, then \(\sum a_n\) converges. This lets us settle messy-sign series by working entirely with positive terms.

The Ratio Test

Let \(L=\displaystyle\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|\). If \(L<1\) the series is absolutely convergent (hence convergent); if \(L>1\) (or \(L=\infty\)) it diverges; if \(L=1\) the test is inconclusive.

The Root Test

Let \(L=\displaystyle\lim_{n\to\infty}\sqrt[n]{|a_n|}\). If \(L<1\) the series converges absolutely; if \(L>1\) (or \(L=\infty\)) it diverges; if \(L=1\) the test is inconclusive. Reach for it when the whole term is raised to the \(n\)th power.

EXAMPLE 1 Ratio Test with a factorial

Test \(\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\frac{n^{3}}{3^{n}}\) for absolute convergence.

Solution

Form the ratio of absolute terms: \(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{(n+1)^{3}}{3^{\,n+1}}\cdot\dfrac{3^{n}}{n^{3}} =\dfrac13\left(\dfrac{n+1}{n}\right)^{3}.\)

As \(n\to\infty\) this tends to \(\dfrac13<1\).

By the Ratio Test the series is absolutely convergent, hence convergent.

EXAMPLE 2 Root Test on an nth power

Test \(\displaystyle\sum_{n=1}^{\infty}\left(\frac{2n+3}{3n+2}\right)^{n}\).

Solution

Take the \(n\)th root: \(\sqrt[n]{a_n}=\dfrac{2n+3}{3n+2}\).

As \(n\to\infty\) this tends to \(\dfrac23<1\).

By the Root Test the series converges.

Rearrangements

An absolutely convergent series may be reordered freely without changing its sum. A conditionally convergent series, by contrast, can be rearranged to sum to any real number — a striking result due to Riemann that shows infinite sums need not behave like finite ones.

SECTION 11.7Strategy for Testing Series

There is no fixed order in which to try the tests. As with integration, the skill is to classify a series by its form and then choose the test that suits that form. The guidelines below summarise the decision process.

How to recognise which test to use

Match the series to the cue, then apply the matching tool:

Looks like \(\sum 1/n^{p}\) → it is a \(p\)-series; converges iff \(p>1\).
Has the form \(\sum ar^{\,n-1}\) → geometric; converges iff \(|r|<1\).
Rational or algebraic in \(n\) → comparison or limit comparison with a \(p\)-series.
You can see at a glance that \(a_n\not\to0\) → Test for Divergence.
Form \(\sum(-1)^{n}b_n\) → Alternating Series Test.
Contains factorials or a constant to the \(n\)th power → Ratio Test.
The whole term is an \(n\)th power → Root Test.
\(a_n=f(n)\) with \(\int f\) easy to evaluate → Integral Test.

EXAMPLE 1 Classifying by form

Which test suits each series? (a) \(\sum\dfrac{n-1}{2n+1}\) (b) \(\sum\dfrac{\sqrt{n^{3}+1}}{3n^{3}+4n^{2}+2}\) (c) \(\sum\dfrac{2^{k}}{k!}\) (d) \(\sum\dfrac{1}{2+3^{n}}\).

Solution

(a) The general term tends to \(\tfrac12\ne0\), so use the Test for Divergence — it diverges.

(b) Algebraic in \(n\); compare with a \(p\)-series using \(b_n\sim n^{-3/2}\) via the Limit Comparison Test.

(c) A factorial appears, so the Ratio Test is natural.

(d) Closely tracks the geometric \(\sum 1/3^{n}\); use the Comparison Test.

Treat §11.7 as a flowchart, not a list. Before computing anything, ask three quick questions in order: Do the terms go to zero? (if not, you're done) — Is it a standard family (geometric / \(p\)-series)? — What structure dominates (factorial, \(n\)th power, alternating sign, rational)? Nine times out of ten the right test is obvious after those three questions.

SECTION 11.8Power Series

A power series is a "polynomial of infinite degree" in a variable \(x\). For each fixed \(x\) it becomes an ordinary numerical series that may converge or diverge, so a power series defines a function on the set of \(x\)-values where it converges.

Power series

A power series centered at \(a\) is \(\sum_{n=0}^{\infty}c_n(x-a)^{n}\), where the constants \(c_n\) are the coefficients. The center \(a=0\) gives the simplest form \(\sum c_n x^{n}\). Every power series converges at its center, since all but the first term vanish when \(x=a\).

Convergence of a power series

For a power series \(\sum c_n(x-a)^{n}\) exactly one of three things happens: it converges only at \(x=a\); it converges for all \(x\); or there is a positive number \(R\) such that it converges for \(|x-a|R\).

Radius and interval of convergence

The number \(R\) above is the radius of convergence (with \(R=0\) for the single-point case and \(R=\infty\) for the everywhere case). The interval of convergence is the full set of \(x\) for which the series converges; at the two endpoints \(x=a\pm R\) anything can happen, so they must be checked separately.

Figure 11.8.1 A power series converges on an interval centered at \(a\) with radius \(R\); the endpoints are decided case by case.

EXAMPLE 1 Finding radius and interval of convergence

Find where \(\displaystyle\sum_{n=1}^{\infty}\frac{(x-3)^{n}}{n}\) converges.

Solution

Ratio Test: \(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{|x-3|^{\,n+1}}{n+1}\cdot\dfrac{n}{|x-3|^{n}} \to|x-3|.\)

Convergence requires \(|x-3|<1\), giving radius \(R=1\) and the open interval \(2

Check endpoints: at \(x=4\) the series is the divergent harmonic series; at \(x=2\) it is the convergent alternating harmonic series. So the interval of convergence is \([2,4)\).

EXAMPLE 2 A Bessel function

Find the domain of the Bessel function of order 0, \(\displaystyle J_0(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{2^{2n}(n!)^{2}}\).

Solution

Apply the Ratio Test to the absolute terms: \(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{x^{2}}{4(n+1)^{2}}\to0\) for every \(x\).

Since the limit is \(0<1\) for all \(x\), the series converges everywhere: the domain is \((-\infty,\infty)\).

SECTION 11.9Representations of Functions as Power Series

Starting from the geometric series, we can manufacture power-series representations of many functions by substitution, and then by differentiating or integrating term by term. This turns hard antiderivatives into easy series.

The fundamental geometric representation

\[ \frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}=1+x+x^{2}+x^{3}+\cdots,\qquad |x|<1. \]

Term-by-term differentiation and integration

If \(f(x)=\sum c_n(x-a)^{n}\) has radius of convergence \(R>0\), then \(f\) is differentiable on \((a-R,a+R)\) and may be differentiated and integrated term by term: \(f'(x)=\sum n\,c_n(x-a)^{n-1}\) and \(\int f\,dx=C+\sum c_n\dfrac{(x-a)^{n+1}}{n+1}\). Both new series keep the same radius \(R\) (endpoints may change).

EXAMPLE 1 Substitution into the geometric series

Represent \(\dfrac{1}{1+x^{2}}\) as a power series.

Solution

Write \(\dfrac{1}{1+x^{2}}=\dfrac{1}{1-(-x^{2})}\) and replace \(x\) by \(-x^{2}\) in the geometric series.

This gives \(\displaystyle\sum_{n=0}^{\infty}(-x^{2})^{n}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n} =1-x^{2}+x^{4}-x^{6}+\cdots,\) valid for \(|x|<1\).

EXAMPLE 2 Integrating to get arctangent

Find a power series for \(\tan^{-1}x\).

Solution

Since \(\dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^{2}}\), integrate the series from Example 1 term by term.

With the constant fixed by \(\tan^{-1}0=0\): \(\displaystyle\tan^{-1}x=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{2n+1} =x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\cdots\) (Gregory's series).

The real payoff of this section shows up in §11.10–11.11 and in integration: functions like \(e^{-x^2}\) and \(\tfrac{1}{1+x^7}\) have no elementary antiderivative, yet expanding them as a series lets you integrate term by term and get any accuracy you want. Whenever you meet "evaluate this integral as an infinite series," this is the toolkit.

SECTION 11.10Taylor and Maclaurin Series

If a function can be written as a power series at all, its coefficients are forced — they must be built from the function's derivatives at the center. This gives the Taylor series, and its special case at the origin, the Maclaurin series.

Taylor series of \(f\) about \(a\)

\[ f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^{n} = f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+\cdots \]

Maclaurin series (Taylor series at \(a=0\))

\[ f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}. \]

Taylor polynomial and remainder

The \(n\)th-degree Taylor polynomial \(T_n(x)\) is the partial sum through the \((x-a)^{n}\) term. Writing \(f(x)=T_n(x)+R_n(x)\), the function equals the sum of its Taylor series on an interval precisely when the remainder satisfies \(\lim_{n\to\infty}R_n(x)=0\) there.

Taylor's Inequality

If \(|f^{(n+1)}(x)|\le M\) for \(|x-a|\le d\), then the remainder of the Taylor series satisfies \(|R_n(x)|\le\dfrac{M}{(n+1)!}\,|x-a|^{\,n+1}\) for \(|x-a|\le d\). This is the standard tool for proving a function equals its series.

EXAMPLE 1 Maclaurin series of the exponential

Find the Maclaurin series of \(f(x)=e^{x}\) and its radius of convergence.

Solution

Every derivative of \(e^{x}\) is \(e^{x}\), so \(f^{(n)}(0)=1\) for all \(n\).

The Maclaurin series is therefore \(\displaystyle\sum_{n=0}^{\infty}\frac{x^{n}}{n!}\).

The Ratio Test gives \(\left|\dfrac{x^{n+1}/(n+1)!}{x^{n}/n!}\right| =\dfrac{|x|}{n+1}\to0\) for every \(x\), so \(R=\infty\).

EXAMPLE 2 Series for sine, then cosine by differentiating

Find the Maclaurin series for \(\sin x\), and deduce the one for \(\cos x\).

Solution

The derivatives of \(\sin x\) cycle \(0,1,0,-1\) at \(x=0\), giving \(\displaystyle\sin x=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!} =x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\cdots\).

Taylor's Inequality with \(M=1\) shows \(R_n\to0\) for all \(x\), so the identity holds everywhere.

Differentiating term by term yields \(\displaystyle\cos x=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!} =1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots\).

The Binomial Series

For any real \(k\) and \(|x|<1\), \(\displaystyle(1+x)^{k}=\sum_{n=0}^{\infty}\binom{k}{n}x^{n} =1+kx+\dfrac{k(k-1)}{2!}x^{2}+\cdots\), where \(\binom{k}{n}\) are the binomial coefficients. When \(k\) is a positive integer the series terminates and reduces to the ordinary Binomial Theorem.

Key Maclaurin series (with radii of convergence)

\[ \frac{1}{1-x}=\sum_{n=0}^{\infty}x^{n}\ (R=1),\qquad e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}\ (R=\infty) \] \[ \sin x=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}\ (R=\infty),\qquad \cos x=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!}\ (R=\infty) \] \[ \tan^{-1}x=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{2n+1}\ (R=1),\qquad (1+x)^{k}=\sum_{n=0}^{\infty}\binom{k}{n}x^{n}\ (R=1) \]

EXAMPLE 3 Integrating a non-elementary function

Evaluate \(\displaystyle\int e^{-x^{2}}\,dx\) as an infinite series.

Solution

Replace \(x\) by \(-x^{2}\) in the series for \(e^{x}\): \(\displaystyle e^{-x^{2}}=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{n!}\).

Integrate term by term: \(\displaystyle\int e^{-x^{2}}\,dx=C+\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)\,n!}.\)

This converges for all \(x\), giving a usable expression even though \(e^{-x^{2}}\) has no elementary antiderivative.

Memorize the six series in the box above — they're the building blocks for almost every series problem you'll meet. The smart move is rarely to compute derivatives from scratch; instead, manipulate a known series by substitution, multiplication, differentiation, or integration (as in Examples 2 and 3). Faster, and far less error-prone.

SECTION 11.11Applications of Taylor Polynomials

Taylor polynomials are how calculators evaluate transcendental functions and how physicists simplify hard formulas: replace a function by the first few terms of its Taylor series, then bound the resulting error.

Polynomial approximation

The \(n\)th-degree Taylor polynomial \(T_n(x)\) approximates \(f\) near the center \(a\), and the approximation improves as \(n\) grows. The quality of the fit is measured by the remainder \(R_n(x)=f(x)-T_n(x)\), which we bound to guarantee accuracy.

Error bound via Taylor's Inequality

\[ |R_n(x)|\le\frac{M}{(n+1)!}\,|x-a|^{\,n+1},\qquad\text{where }|f^{(n+1)}(x)|\le M. \]

Figure 11.11.1 Successive Taylor polynomials of \(\sin x\) hug the curve over a widening interval; each added term extends the range of good agreement.

EXAMPLE 1 Bounding the error of a sine approximation

Estimate the maximum error in \(\sin x\approx x-\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}\) for \(|x|\le0.3\), and use it to find \(\sin 12^{\circ}\).

Solution

The series is alternating, so the error is at most the first omitted term \(\dfrac{|x|^{7}}{7!}\le\dfrac{(0.3)^{7}}{5040}\approx4.3\times10^{-8}\).

Convert \(12^{\circ}=\dfrac{\pi}{15}\) radians and substitute, giving \(\sin 12^{\circ}\approx0.207912\) — correct to six decimals.

EXAMPLE 2 Special relativity: recovering Newtonian energy

The relativistic kinetic energy is \(K=m_0c^{2}\!\left[(1-v^{2}/c^{2})^{-1/2}-1\right]\). Show that for \(v\ll c\) it reduces to the classical \(K\approx\tfrac12 m_0v^{2}\).

Solution

Expand \((1-x)^{-1/2}\) by the binomial series with \(x=v^{2}/c^{2}\): \(1+\tfrac12 x+\tfrac38 x^{2}+\cdots\).

Then \(K=m_0c^{2}\!\left[\tfrac12\dfrac{v^{2}}{c^{2}}+\tfrac38\dfrac{v^{4}}{c^{4}}+\cdots\right].\)

When \(v\ll c\) the higher terms are negligible, leaving \(K\approx\tfrac12 m_0v^{2}\) — exactly the Newtonian formula.

This section is where the whole chapter pays off for engineers. The pattern in Example 2 — expand, keep the leading term(s), discard the rest with an error bound — is how relativistic corrections, small-angle pendulum periods, lens (Gaussian optics) equations, and earth-curvature highway corrections are all derived. When a formula is intractable, a degree-1 or degree-2 Taylor polynomial usually reveals the physics hiding inside it.

End of Chapter 11. Section exercises, review questions, and chapter review problems are omitted from this reference by design.