Unit 10: Infinite Sequences and Series

Sequences

Definition of a Sequence and Applying Limits to Understand Convergence

• A sequence is an ordered list of numbers, typically written as \( \{a_n\} \), where each term is defined by a formula or rule. To determine whether a sequence converges, evaluate \( \lim_{n \to \infty} a_n \) using algebraic or calculus-based methods. If the limit exists and is finite, the sequence converges to that value; if it does not exist or is infinite, the sequence diverges.
• Step-by-step:
  (1) Identify the general term \( a_n \).
  (2) Simplify the expression for large \( n \) using factoring or dividing by the highest power of \( n \).
  (3) Apply limit laws or L’Hôpital’s Rule if necessary. Example: \( a_n = \frac{3n^2 + 4}{2n^2 - 1} \) → divide by \( n^2 \) to get \( \frac{3 + 4/n^2}{2 - 1/n^2} \) → limit is \( \frac{3}{2} \), so it converges.
• Connects to series: Since the convergence of a series \(\sum a_n\) requires \( a_n \to 0 \), understanding how to find limits of sequences is the first building block in series analysis. Incorrectly computing these limits can lead to false conclusions about a series’ behavior.
• Example: \( a_n = \frac{\sin n}{n} \) → \( |\sin n| \le 1 \) and \( \frac{1}{n} \to 0 \) by the Squeeze Theorem, so \( a_n \to 0 \) and the sequence converges to 0. This technique is essential for sequences with bounded oscillation divided by a growing denominator.
• Common mistake: assuming a sequence converges just because the numerator grows slower than the denominator without verifying the limit formally. Always provide a clear step-by-step limit calculation or inequality proof to avoid this.

Monotonic Sequences and the Monotone Convergence Theorem

• A sequence is monotonic if it is either always increasing (\( a_{n+1} \ge a_n \)) or always decreasing (\( a_{n+1} \le a_n \)). The Monotone Convergence Theorem states that if a sequence is monotonic and bounded, it must converge. This theorem is useful for proving convergence when direct limit evaluation is difficult.
• Step-by-step:
  (1) Prove monotonicity by checking \( a_{n+1} - a_n \) or \( \frac{a_{n+1}}{a_n} \).
  (2) Establish a bound by comparing to a constant.
  (3) Conclude convergence from the theorem. Example: \( a_n = 1 - \frac{1}{n} \) → \( a_{n+1} - a_n = \frac{1}{n} - \frac{1}{n+1} > 0 \) so increasing, and clearly \( a_n < 1 \), so it converges.
• Connection to series: Monotonic sequences appear in the Alternating Series Test’s remainder estimation and in the behavior of partial sums for certain series. Recognizing monotonicity can give insight into the stability of sums.
• Example: \( a_n = \sqrt{2 + a_{n-1}} \) with \( a_1 = 1 \). Prove \( a_n \) is increasing by squaring carefully, then show \( a_n \le 2 \) as an upper bound. The limit \( L \) must satisfy \( L = \sqrt{2 + L} \) → \( L = 2 \).
• Common mistake: assuming a sequence is monotonic by looking at the first few terms without proof. Always verify the property algebraically or via induction.

Recursive Sequences and Fixed Points

• A recursive sequence defines each term from previous terms, such as \( a_{n+1} = f(a_n) \). Convergence often involves finding a fixed point \( L \) where \( L = f(L) \). Stability depends on whether values get closer to \( L \) as \( n \) increases.
• Step-by-step: (1) Identify the recursion rule. (2) Prove monotonicity and boundedness if possible. (3) Solve \( L = f(L) \) to find potential limits. (4) Verify that the sequence actually approaches \( L \) by testing a few terms and using inequalities. Example: \( a_{n+1} = \frac{1}{2}(a_n + \frac{5}{a_n}) \) converges to \( \sqrt{5} \) (Babylonian method).
• Connection to calculus: Many numerical methods (like Newton’s Method) generate recursive sequences, so understanding fixed points ties into later BC concepts such as root finding and error estimation.
• Example: \( a_1 = 2, \; a_{n+1} = \frac{a_n + 3}{a_n + 1} \). Show it’s bounded and decreasing, then solve \( L = \frac{L + 3}{L + 1} \) → \( L^2 + L = L + 3 \) → \( L^2 = 3 \) → \( L = \sqrt{3} \).
• Common mistake: assuming the fixed point is the limit without proving stability. A fixed point can be unstable, causing divergence if the starting value is far enough away.

Recognizing Divergence Patterns in Sequences

• Some sequences diverge by growing without bound (\( a_n \to \infty \)), oscillating indefinitely, or failing to approach a single value. Recognizing these quickly can save time on exams. If \( \lim_{n\to\infty} a_n \) does not exist or is infinite, the sequence diverges.
• Step-by-step: (1) Check for polynomial or exponential growth. (2) If oscillatory, examine whether magnitude decreases (e.g., \( \frac{(-1)^n}{n} \) converges to 0). (3) Look for repeating patterns in trigonometric sequences. Example: \( a_n = (-1)^n \) oscillates between -1 and 1 → diverges.
• Connection to series: If a sequence’s terms don’t approach 0, its series will fail the n-th term test for convergence. Divergence recognition is thus the first checkpoint before applying any series test.
• Example: \( a_n = n\sin\left(\frac{\pi n}{2}\right) \) alternates between large positive, large negative, and zero values → diverges by unbounded oscillation.
• Common mistake: thinking oscillation always means divergence. If oscillation occurs but amplitude decreases to zero, the sequence can converge (e.g., \( \frac{\cos n}{n^2} \)).

Common Sequence Limit Techniques

• Divide by the highest power of \( n \) for rational expressions. Use L’Hôpital’s Rule for indeterminate forms when treating \( a_n \) as \( f(n) \). Example: \( \lim_{n\to\infty} \frac{\ln n}{n} = 0 \) since \( n \) grows faster than \( \ln n \).
• Apply the Squeeze Theorem when the sequence is bounded above and below by sequences with the same limit. Example: \( 0 \le \frac{\sin n}{n} \le \frac{1}{n} \) → both bounds → 0.
• For exponential terms, compare growth rates: exponentials outpace polynomials, which outpace logarithms. This hierarchy is key for both sequences and series tests like the Ratio Test.
• For factorial terms, use Stirling’s approximation or compare \( n! \) to \( n^n \) growth rates. Example: \( \frac{n^2}{n!} \to 0 \) because factorial growth dominates.
• Common mistake: applying continuous methods without justification. Always confirm that \( f(x) \) is continuous for \( x\ge N \) before replacing \( n \) with \( x \) to compute limits.

Series Basics

Definition of a Series and Partial Sums

• An infinite series is the sum of terms from a sequence: \( \sum_{n=1}^{\infty} a_n \). The key idea is that we don’t add infinitely many numbers all at once; instead, we form partial sums \( S_N = \sum_{n=1}^N a_n \) and see if \( S_N \) approaches a finite limit as \( N \to \infty \).
• Step-by-step:
  (1) Identify the term \( a_n \).
  (2) Write out the first few partial sums.
  (3) If possible, find a formula for \( S_N \).
  (4) Take \( \lim_{N \to \infty} S_N \). If the limit exists and is finite, the series converges; otherwise, it diverges.
• Connection to sequences: The sequence \(\{S_N\}\) is the sequence of partial sums. A series converges exactly when this sequence converges, so all series problems boil down to sequence limit problems.
• Example: \( a_n = \frac{1}{n(n+1)} \) → partial sums telescope to \( S_N = 1 - \frac{1}{N+1} \), and \(\lim_{N\to\infty} S_N = 1\), so the series converges.
• Common mistake: confusing the limit of \( a_n \) with the limit of \( S_N \). Even if \( a_n \to 0 \), the series can still diverge (harmonic series). Always analyze \( S_N \) for convergence.

Applying Limits to Understand Convergence of Infinite Series

• The first test to apply to any series is the n-th term test for divergence: if \( \lim_{n \to \infty} a_n \neq 0 \), then \( \sum a_n \) diverges. This is a necessary condition for convergence but not sufficient.
• Step-by-step:
  (1) Compute \( \lim_{n \to \infty} a_n \).
  (2) If the limit is nonzero or does not exist, conclude divergence immediately.
  (3) If the limit is zero, proceed to another convergence test—never stop here.
• Example: \( a_n = \frac{n+1}{n} \to 1 \neq 0 \) → series diverges without any further work. Example: \( a_n = \frac{1}{n} \to 0 \) but series diverges (harmonic), so another test is needed.
• Connection to sequences: This test uses the exact same limit skills as sequence convergence, reinforcing why mastery of sequence limits is essential before moving to series.
• Common mistake: thinking that if \(\lim a_n = 0\), the series converges. This leads to wrong answers for famous counterexamples like the harmonic series.

Types of Series: Geometric, Harmonic, and \( p \)-Series

• A geometric series has the form \( \sum_{n=0}^{\infty} ar^n \) with ratio \( r \). It converges if \( |r| < 1 \) and diverges otherwise. Sum formula (when convergent): \( S = \frac{a}{1-r} \).
• The harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \) diverges despite terms going to zero. It is the \( p \)-series case with \( p = 1 \), which is the threshold between convergence and divergence.
• A \( p \)-series is \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). It converges if \( p > 1 \) and diverges if \( 0 < p \le 1 \). This rule is widely used for comparison tests.
• Example: \( \sum_{n=0}^{\infty} 3\left(\frac{1}{2}\right)^n \) is geometric with \( r = \frac{1}{2} \) → sum is \( \frac{3}{1 - 1/2} = 6 \). Example: \( \sum \frac{1}{n^2} \) converges (\( p = 2 > 1 \)).
• Common mistake: forgetting to check \( |r| < 1 \) before applying the geometric sum formula or misidentifying \( p \) in a \( p \)-series with extra terms in the denominator.

Telescoping Series and Partial Fraction Decomposition

• A telescoping series collapses because consecutive terms cancel out in partial sums. To reveal this structure, rewrite \( a_n \) as a difference of fractions (partial fractions) or other expressions.
• Step-by-step:
  (1) Decompose \( a_n \) into \( b_n - b_{n+1} \).
  (2) Write out first few partial sums to observe cancellation.
  (3) Take \( \lim_{N \to \infty} S_N \) using the leftover terms.
• Example: \( a_n = \frac{1}{n} - \frac{1}{n+1} \) → partial sum \( S_N = 1 - \frac{1}{N+1} \) → limit is 1. Another: \( \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \).
• Connection to convergence: Telescoping can show exact sums without estimating. Often used when \( a_n \) comes from a rational function that simplifies.
• Common mistake: claiming telescoping without showing the partial sums. Always demonstrate the actual cancellation pattern before concluding.

Convergence Tests

n-th Term Test for Divergence

• If \( \lim_{n \to \infty} a_n \neq 0 \) or the limit does not exist, the series \( \sum a_n \) diverges. This is always the first check to perform before applying other tests.
• Step-by-step: (1) Find \( \lim_{n \to \infty} a_n \). (2) If the limit is not zero or does not exist, conclude divergence immediately. (3) If the limit is zero, continue to another test.
• Example: \( a_n = \frac{n+2}{n} \) has limit \( 1 \), so \( \sum a_n \) diverges.
• Example: \( a_n = \frac{1}{n} \) has limit \( 0 \), so the test is inconclusive and another method is needed.
• Always apply this test before spending time on more complex tests to quickly identify divergence.

Direct Comparison Test (DCT)

• Use when you can compare \( a_n \) directly to a known convergent or divergent series with positive terms. If \( 0 \le a_n \le b_n \) for all \( n \) and \( \sum b_n \) converges, then \( \sum a_n \) converges. If \( 0 \le b_n \le a_n \) for all \( n \) and \( \sum b_n \) diverges, then \( \sum a_n \) diverges.
• Step-by-step:
  (1) Identify a simpler benchmark series \( b_n \).
  (2) Write the inequality between \( a_n \) and \( b_n \).
  (3) State the known behavior of \( \sum b_n \). (4) Conclude the behavior of \( \sum a_n \).
• Example: \( a_n = \frac{1}{n^2 + n} \le \frac{1}{n^2} \) and \( \sum \frac{1}{n^2} \) converges (\( p = 2 > 1 \)), so \( \sum a_n \) converges.
• Example: \( a_n = \frac{1}{n} \ge \frac{1}{2n} \) and \( \sum \frac{1}{2n} \) diverges (harmonic), so \( \sum a_n \) diverges.
• This test works best when terms can be cleanly compared to a p-series or geometric series.

Limit Comparison Test (LCT)

• Use when the Direct Comparison Test is hard to apply but the terms behave similarly to a known series. If \( \lim_{n \to \infty} \frac{a_n}{b_n} = L \) where \( L \) is a finite positive number, then \( \sum a_n \) and \( \sum b_n \) either both converge or both diverge.
• Step-by-step:
  (1) Choose \( b_n \) based on the dominant term in \( a_n \).
  (2) Compute \( \lim_{n \to \infty} \frac{a_n}{b_n} \).
  (3) If \( L > 0 \) and finite, the two series have the same behavior.
• Example: \( a_n = \frac{n^2 + 3}{5n^2 - 1} \) with \( b_n = \frac{1}{n^2} \) → ratio limit \( = \frac{1}{5} \), and since \( \sum \frac{1}{n^2} \) converges, \( \sum a_n \) converges.
• Example: \( a_n = \frac{n}{n^2 + 4} \) with \( b_n = \frac{1}{n} \) → ratio limit \( = 1 \), and since \( \sum \frac{1}{n} \) diverges, \( \sum a_n \) diverges.
• This is most useful for rational expressions and terms involving roots where direct inequality comparisons are difficult.

Integral Test

• Applies when \( a_n = f(n) \) where \( f \) is positive, continuous, and decreasing for \( x \ge N \). The series \( \sum a_n \) converges if and only if \( \int_N^\infty f(x) \, dx \) converges.
• Step-by-step: (1) Verify that \( f(x) \) meets the criteria. (2) Evaluate \( \int_N^\infty f(x) \, dx \) as an improper integral. (3) Conclude that the series converges if the integral is finite and diverges if it is infinite.
• Example: \( a_n = \frac{1}{n^p} \) → \( \int_1^\infty x^{-p} dx \) converges if \( p > 1 \) and diverges if \( p \le 1 \).
• Example: \( a_n = \frac{1}{n \ln n} \) → \( \int_2^\infty \frac{1}{x \ln x} dx = \ln(\ln x) \big|_2^\infty \) diverges, so the series diverges.
• This test also provides a way to estimate the remainder when approximating a sum by a partial sum.

Alternating Series Test (Leibniz Test)

• An alternating series has terms that switch signs, typically written as \( \sum (-1)^{n} b_n \) or \( \sum (-1)^{n+1} b_n \) with \( b_n > 0 \). The series converges if \( b_n \) is decreasing and \( \lim_{n \to \infty} b_n = 0 \).
• Step-by-step:
  (1) Identify \( b_n \).
  (2) Verify \( b_{n+1} \le b_n \) for all large \( n \).
  (3) Check that \( b_n \to 0 \). If both hold, the series converges.
• Example: \( \sum (-1)^{n+1} \frac{1}{n} \) satisfies decreasing \( b_n = \frac{1}{n} \) and \( b_n \to 0 \), so it converges.
• Example: \( \sum (-1)^n \frac{n}{n+1} \) fails because \( b_n \to 1 \neq 0 \), so it diverges.
• The Alternating Series Error Bound states that the error in approximating by the \( N \)-th partial sum is less than or equal to \( b_{N+1} \).

Absolute and Conditional Convergence

• A series \( \sum a_n \) converges absolutely if \( \sum |a_n| \) converges. Absolute convergence guarantees convergence of \( \sum a_n \).
• If \( \sum a_n \) converges but \( \sum |a_n| \) diverges, then the series converges conditionally. This often happens with alternating series that fail absolute convergence.
• Example: \( \sum \frac{(-1)^n}{n} \) converges conditionally because \( \sum \frac{1}{n} \) diverges.
• Example: \( \sum \frac{(-1)^n}{n^2} \) converges absolutely because \( \sum \frac{1}{n^2} \) converges.
• Absolute convergence is stronger and should be checked first before considering conditional convergence.

Ratio Test and Root Test

• Ratio Test: Compute \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). If \( L < 1 \), the series converges absolutely. If \( L > 1 \) or is infinite, the series diverges. If \( L = 1 \), the test is inconclusive.
• Root Test: Compute \( L = \lim_{n \to \infty} \sqrt[n]{|a_n|} \). Same conclusions as the Ratio Test: \( L < 1 \) → converges absolutely, \( L > 1 \) → diverges, \( L = 1 \) → inconclusive.
• Step-by-step for Ratio Test:
  (1) Write \( \frac{a_{n+1}}{a_n} \).
  (2) Simplify completely.
  (3) Take the limit as \( n \to \infty \).
  (4) Compare to 1 for conclusion.
• Step-by-step for Root Test:
  (1) Take the n-th root of \( |a_n| \).
  (2) Simplify expression.
  (3) Take the limit.
  (4) Compare to 1 for conclusion.
• Example Ratio Test: \( a_n = \frac{n!}{n^n} \) → \( L = \frac{1}{e} < 1 \), so converges absolutely. Example Root Test: \( a_n = \left( \frac{n}{n+1} \right)^n \) → \( L = \frac{1}{e} < 1 \), so converges absolutely.

Approximating Sums and Error Bounds

Approximating the Sum of a Convergent Series

• To approximate the sum of a convergent series \( \sum a_n \), choose a partial sum \( S_N = \sum_{n=1}^N a_n \) that adds the first \( N \) terms. This partial sum is your approximation to the exact infinite sum. The difference between \( S_N \) and the exact sum is called the remainder \( R_N \), and we want this remainder to be as small as needed for the problem.
• Step-by-step:
  (1) Identify that the series converges using an appropriate test.
  (2) Decide on the number of terms \( N \) you will include based on the desired accuracy.
  (3) Compute \( S_N \) directly by summing those terms.
  (4) Use an error bound method to confirm that \( |R_N| \) is within tolerance.
  (5) State the approximation as \( S \approx S_N \) with an error less than the bound.
• Example: To approximate \( \sum_{n=1}^\infty \frac{1}{n^2} \) to within \( 0.01 \), use the integral test remainder bound \( R_N \le \int_{N}^\infty \frac{1}{x^2} dx = \frac{1}{N} \). Solve \( \frac{1}{N} \le 0.01 \) → \( N \ge 100 \). Compute \( S_{100} \) and guarantee that the sum is within 0.01 of the true value.
• This method is required when the infinite sum cannot be computed exactly. The partial sum gives a practical numerical value, while the error bound ensures the answer is accurate to the required number of decimal places.
• Always explicitly show the steps: proving convergence, calculating the partial sum, and bounding the remainder. Skipping any of these leaves the approximation unsupported.

Integral Test Remainder Estimate

• For a positive, decreasing series where the integral test applies, the remainder after \( N \) terms satisfies \( \int_{N+1}^\infty f(x) \, dx \le R_N \le \int_{N}^\infty f(x) \, dx \) where \( f(n) = a_n \). This gives an upper and lower bound for the error made when approximating with \( S_N \).
• Step-by-step:
  (1) Verify \( f(x) \) is positive, continuous, and decreasing for \( x \ge N \).
  (2) Write the upper bound \( R_N \le \int_{N}^\infty f(x) dx \).
  (3) Solve the inequality to find \( N \) needed for the desired error.
  (4) Compute \( S_N \) with that \( N \).
  (5) State the final approximation and guaranteed error.
• Example: For \( \sum_{n=1}^\infty \frac{1}{n^p} \) with \( p > 1 \), \( R_N \le \int_{N}^\infty x^{-p} dx = \frac{1}{(p-1)N^{p-1}} \). To get error below \( 0.001 \) when \( p = 2 \), solve \( \frac{1}{N} \le 0.001 \) → \( N \ge 1000 \).
• The integral bounds are tight enough for accurate estimates and work even for slowly converging series when other methods are difficult to apply.
• Be precise: compute both the bound and the partial sum so the approximation is fully justified. The bound alone is not enough to give the approximation.

Alternating Series Error Bound

• For an alternating series that meets the Alternating Series Test conditions, the error when stopping at \( N \) terms is less than or equal to the absolute value of the first omitted term: \( |R_N| \le |a_{N+1}| \).
• Step-by-step:
  (1) Verify \( b_n \) is decreasing and \( b_n \to 0 \) where \( a_n = (-1)^n b_n \).
  (2) Choose \( N \) so that \( b_{N+1} \) is less than the desired error tolerance.
  (3) Compute \( S_N \) as the approximation.
  (4) State \( S \approx S_N \) with \( |R_N| \le b_{N+1} \).
• Example: For \( \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \), to get error \( \le 0.001 \), solve \( \frac{1}{N+1} \le 0.001 \) → \( N+1 \ge 1000 \) → \( N \ge 999 \). Compute \( S_{999} \) for the approximation.
• This bound is exact in the sense that it uses the next term size directly, which is simple and avoids integration.
• Always check decreasing terms and limit to zero before applying; if these fail, the bound is invalid.

Taylor Series Error Bound (Lagrange Remainder)

• When approximating a function \( f(x) \) with a Taylor polynomial \( P_n(x) \), the Lagrange form of the remainder is \( R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} \) for some \( c \) between \( a \) and \( x \). This gives a way to bound the error by finding a maximum possible value of \( |f^{(n+1)}(t)| \) on that interval.
• Step-by-step:
  (1) Write the Taylor polynomial \( P_n(x) \) for \( f(x) \) centered at \( a \).
  (2) Find \( f^{(n+1)}(x) \) and bound its absolute value on the interval between \( a \) and \( x \).
  (3) Plug the bound into \( |R_n(x)| \le \frac{M}{(n+1)!}|x-a|^{n+1} \).
  (4) Ensure the bound meets the required accuracy.
• Example: Approximate \( e^x \) at \( x = 0.5 \) using \( n = 3 \) centered at \( a = 0 \). Here \( f^{(4)}(x) = e^x \) and on \( [0, 0.5] \), \( e^x \le e^{0.5} \). So \( M = e^{0.5} \), and \( |R_3(0.5)| \le \frac{e^{0.5}}{4!}(0.5)^4 \).
• This method ensures the polynomial approximation is within a known error margin, which is essential for accuracy in applications and AP free-response questions.
• Always choose \( M \) carefully by taking the largest value of the derivative on the interval, not just at the endpoints.

Power Series

Definition and General Form of a Power Series

• A power series is an infinite series of the form \( \sum_{n=0}^{\infty} c_n (x-a)^n \), where \( c_n \) are constants and \( a \) is the center. The variable \( x \) is the input, and the series becomes a function that outputs a sum depending on \( x \). Convergence depends on how far \( x \) is from \( a \).
• Step-by-step:
  (1) Identify \( c_n \) and the center \( a \).
  (2) Recognize that for different \( x \) values, the series may converge or diverge.
  (3) Determine the set of \( x \) values where it converges; this set is called the interval of convergence.
  (4) Use convergence tests, typically the Ratio or Root Test, to find where convergence happens.
• Example: \( \sum_{n=0}^{\infty} \frac{(x-2)^n}{n!} \) is a power series centered at \( a = 2 \) with coefficients \( c_n = \frac{1}{n!} \). This series converges for all real \( x \) because factorial growth in the denominator ensures convergence everywhere.
• Power series are important because they act like polynomials of infinite degree and can approximate functions extremely accurately within their interval of convergence.
• When the center is \( a = 0 \), the series is called a Maclaurin series. Otherwise, it is a Taylor series centered at \( a \).

Determining the Radius of Convergence

• The radius of convergence \( R \) describes the distance from the center \( a \) within which the series converges absolutely. To find \( R \), apply the Ratio Test or Root Test to the general term \( c_n (x-a)^n \).
• Step-by-step using Ratio Test: (1) Compute \( L = \lim_{n \to \infty} \left| \frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n} \right| \). (2) Simplify to \( L = |x-a| \cdot \lim_{n \to \infty} \left| \frac{c_{n+1}}{c_n} \right| \). (3) Require \( L < 1 \) for convergence. (4) Solve \( |x-a| < R \) to find the radius.
• Example: For \( \sum_{n=1}^\infty \frac{(x-3)^n}{n} \), \( L = |x-3| \lim_{n \to \infty} \frac{n}{n+1} = |x-3| \). Convergence requires \( |x-3| < 1 \), so \( R = 1 \).
• The Root Test can be used similarly: \( L = \lim_{n\to\infty} \sqrt[n]{|c_n||x-a|^n} = |x-a| \lim_{n\to\infty} \sqrt[n]{|c_n|} \), and convergence requires \( L < 1 \).
• The radius \( R \) tells you the “open” part of the interval where the series converges absolutely; endpoints require separate testing.

Determining the Interval of Convergence

• The interval of convergence is the set of all \( x \) values for which the series converges. It always has the form \( (a-R, a+R) \) or some variation with endpoints included depending on endpoint behavior.
• Step-by-step:
  (1) Find \( R \) using Ratio or Root Test.
  (2) Write the open interval \( (a-R, a+R) \).
  (3) Test \( x = a-R \) and \( x = a+R \) separately using appropriate convergence tests (often Alternating Series Test, p-series comparison, or direct evaluation).
  (4) Decide whether to include each endpoint based on convergence or divergence there.
• Example: For \( \sum_{n=1}^\infty \frac{(x+2)^n}{n} \), \( R = 1 \) from Ratio Test. Open interval: \( (-3, -1) \). At \( x = -3 \), terms are \( \frac{(-1)^n}{n} \) which converges by Alternating Series Test → include \( -3 \). At \( x = -1 \), terms are \( \frac{1}{n} \) which diverges (harmonic) → exclude \( -1 \). Final interval: \( [-3, -1) \).
• Always test both endpoints independently, since convergence inside the radius does not guarantee convergence at the boundary.
• The interval of convergence must be written carefully, with correct bracket notation showing inclusion or exclusion of endpoints.

Convergence at Endpoints

• At \( x = a \pm R \), the Ratio and Root Tests give \( L = 1 \), so they are inconclusive. You must substitute the endpoint into the series and analyze it as a separate series using the tests from convergence theory.
• Step-by-step:
  (1) Plug \( x = a-R \) into \( \sum c_n (x-a)^n \) to get a numeric series.
  (2) Apply tests like the p-series test, Alternating Series Test, or Comparison Tests.
  (3) Record whether the series converges or diverges at that endpoint.
  (4) Repeat for \( x = a+R \).
• Example: For \( \sum_{n=1}^\infty \frac{(x-4)^n}{n} \) with \( R = 2 \), at \( x = 6 \), the series is \( \sum \frac{2^n}{n} \) which diverges by the Ratio Test. At \( x = 2 \), the series is \( \sum \frac{(-2)^n}{n} \) which converges conditionally by Alternating Series Test.
• This process ensures the interval of convergence is complete and accurate, avoiding the common mistake of ignoring endpoints.
• Endpoint analysis is required on AP exams for full credit when determining intervals of convergence.

Taylor and Maclaurin Polynomials

Definition and Purpose of Taylor and Maclaurin Polynomials

• A Taylor polynomial of degree \( n \) for a function \( f(x) \) centered at \( a \) is a polynomial that matches the value of \( f \) and its first \( n \) derivatives at \( x = a \). Its general form is \( P_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n \).
• When the center \( a = 0 \), the polynomial is called a Maclaurin polynomial, which is a special case of a Taylor polynomial centered at zero. The formula becomes \( P_n(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots + \frac{f^{(n)}(0)}{n!}x^n \).
• Step-by-step to construct: (1) Identify the function \( f(x) \) and center \( a \). (2) Compute derivatives \( f'(x), f''(x), \dots, f^{(n)}(x) \). (3) Evaluate each derivative at \( x = a \). (4) Substitute into the Taylor formula up to degree \( n \).
• Example: For \( f(x) = e^x \) centered at \( a = 0 \) with \( n = 3 \), all derivatives are \( e^x \), so \( f^{(k)}(0) = 1 \) for all \( k \). The Maclaurin polynomial is \( P_3(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} \).
• Taylor and Maclaurin polynomials are used to approximate functions near the center \( a \), and the approximation becomes more accurate as \( n \) increases or \( x \) is closer to \( a \).

Approximating Functions with Taylor Polynomials

• To approximate \( f(x) \) near \( x = a \), compute \( P_n(x) \) and use it as the approximation \( f(x) \approx P_n(x) \). This is especially useful when evaluating \( f(x) \) exactly is difficult or impossible.
• Step-by-step:
  (1) Construct \( P_n(x) \) using the method above.
  (2) Choose \( n \) based on the desired accuracy or error bound.
  (3) Evaluate \( P_n(x) \) at the target \( x \).
  (4) If required, compute the error bound to confirm the accuracy.
• Example: Approximate \( \cos(0.1) \) with \( n = 2 \) centered at \( a = 0 \). Derivatives alternate between \( \cos x \) and \( -\sin x \), so \( f(0) = 1 \), \( f'(0) = 0 \), \( f''(0) = -1 \). Polynomial: \( P_2(x) = 1 - \frac{x^2}{2} \). Approximation: \( 1 - \frac{(0.1)^2}{2} = 0.995 \).
• Higher-degree polynomials generally give more accurate approximations within a given interval around \( a \). However, accuracy can worsen if \( x \) is far from \( a \), even for large \( n \).
• This technique connects directly to Taylor series, which are the infinite-degree version of Taylor polynomials.

Error Estimation with Lagrange Remainder

• The Lagrange Remainder formula gives a bound for the error when approximating \( f(x) \) by \( P_n(x) \): \( R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} \) for some \( c \) between \( a \) and \( x \). The magnitude \( |R_n(x)| \) is bounded by replacing \( |f^{(n+1)}(c)| \) with a maximum \( M \) on that interval.
• Step-by-step:
  (1) Determine \( f^{(n+1)}(x) \).
  (2) Find the maximum value \( M \) of \( |f^{(n+1)}(x)| \) for \( x \) between \( a \) and the target \( x \).
  (3) Substitute \( M \), \( n \), \( a \), and \( x \) into \( \frac{M}{(n+1)!}|x-a|^{n+1} \) to get an error bound.
  (4) Compare the bound to the required tolerance to confirm adequacy.
• Example: Approximate \( e^x \) at \( x = 0.5 \) with \( n = 2 \) centered at \( a = 0 \). The third derivative is \( e^x \), and on \( [0,0.5] \), \( e^x \le e^{0.5} \). Bound: \( |R_2(0.5)| \le \frac{e^{0.5}}{3!}(0.5)^3 \).
• This method is necessary to prove accuracy rather than just assuming it. AP exam problems often require stating the error bound explicitly.
• Calculating \( M \) correctly ensures the bound is valid; use the maximum derivative value on the entire interval, not just at a point.

Taylor and Maclaurin Series

Definition of a Taylor Series

• A Taylor series is the infinite sum obtained by extending a Taylor polynomial to infinitely many terms. For a function \( f(x) \) centered at \( a \), the Taylor series is \( \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n \). This series can represent \( f(x) \) exactly for all \( x \) within its interval of convergence.
• When \( a = 0 \), the series becomes a Maclaurin series: \( \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n \). This is a special case where the series is centered at zero, making many formulas simpler.
• Step-by-step to form a Taylor series: (1) Compute derivatives \( f'(x), f''(x), \dots \). (2) Evaluate each at \( x = a \). (3) Substitute into the general term \( \frac{f^{(n)}(a)}{n!}(x-a)^n \). (4) Write the infinite sum starting at \( n = 0 \).
• Example: For \( f(x) = e^x \) centered at \( a = 0 \), \( f^{(n)}(0) = 1 \) for all \( n \), so the Maclaurin series is \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \).
• Unlike polynomials, which are finite sums, a Taylor series can capture the exact behavior of a function on its interval of convergence if the remainder term goes to zero.

Common Maclaurin Series

• Certain functions have well-known Maclaurin series that are worth memorizing for quick use on exams:
• \( e^x = \sum_{n=0}^\infty \frac{x^n}{n!} \), valid for all \( x \).
• \( \sin x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1} \), valid for all \( x \).
• \( \cos x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n} \), valid for all \( x \).
• \( \frac{1}{1-x} = \sum_{n=0}^\infty x^n \), valid for \( |x| < 1 \).
• \( \ln(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n \), valid for \( -1 < x \le 1 \).

Generating New Series from Known Series

• You can create new Taylor or Maclaurin series by substituting, differentiating, or integrating known series. These operations preserve the radius of convergence (except possibly at endpoints).
• Step-by-step:
  (1) Start with a known series, like \( \frac{1}{1-x} = \sum x^n \).
  (2) Replace \( x \) with another expression, e.g., \( 2x^2 \) to get \( \frac{1}{1-2x^2} = \sum (2x^2)^n \).
  (3) Differentiate or integrate term-by-term if needed to match your target function.
• Example: To find the Maclaurin series for \( \frac{x}{1-x} \), start with \( \frac{1}{1-x} = \sum x^n \) and multiply the whole series by \( x \) to get \( \sum x^{n+1} \).
• Example: To find \( \arctan x \), start with \( \frac{1}{1+x^2} = \sum (-1)^n x^{2n} \) for \( |x| < 1 \), then integrate term-by-term to get \( \arctan x = \sum \frac{(-1)^n}{2n+1}x^{2n+1} \).
• These transformations are standard exam techniques and save time compared to recomputing derivatives from scratch.

Determining Intervals of Convergence for Taylor/Maclaurin Series

• Just like power series, a Taylor or Maclaurin series converges on an interval determined by its radius of convergence \( R \) and possible endpoint behavior. The interval must be explicitly determined for complete accuracy.
• Step-by-step:
  (1) Apply the Ratio or Root Test to the general term of the series to find \( R \).
  (2) Write the open interval \( (a-R, a+R) \) where convergence is guaranteed.
  (3) Test each endpoint individually using standard convergence tests.
  (4) State the full interval using brackets for included endpoints and parentheses for excluded ones.
• Example: For \( \ln(1+x) \), the Ratio Test gives \( R = 1 \). At \( x = 1 \), series becomes \( \sum \frac{(-1)^{n+1}}{n} \) which converges (alternating harmonic). At \( x = -1 \), series becomes \( \sum \frac{(-1)^{n+1}}{n}(-1)^n = \sum \frac{-1}{n} \) which diverges. Interval: \( (-1, 1] \).
• Without testing endpoints, you cannot determine the full convergence set, which can lead to incomplete answers.
• This process ensures you can use the series safely within the range where it truly represents the function.

Practice Problems

Problem 1 (Representing a Function as a Taylor Series and Finding Its Interval)

• Question: Find the Maclaurin series for \( \ln(1+3x) \). Determine the radius and interval of convergence, and write the first four nonzero terms explicitly.
• Step 1 (series construction): Start from the known Maclaurin series \( \ln(1+u)=\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}u^{\,n} \) for \( -1
• Step 2 (radius of convergence): The ratio test on the general term \( \frac{(-1)^{n+1}}{n}(3x)^n \) gives \( \lim_{n\to\infty}\big|\frac{a_{n+1}}{a_n}\big|=|3x| \). Convergence requires \( |3x|<1 \), so the radius is \( R=\tfrac{1}{3} \). The open interval is \( (-\tfrac{1}{3},\tfrac{1}{3}) \) before endpoint testing.
• Step 3 (endpoint analysis): At \( x=\tfrac{1}{3} \), the series becomes \( \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} \), which converges (alternating harmonic) and sums to \( \ln 2 \). At \( x=-\tfrac{1}{3} \), the series becomes \( \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}(-1)^n=-\sum_{n=1}^{\infty}\frac{1}{n} \), which diverges. Therefore the interval of convergence is \( \left(-\tfrac{1}{3},\,\tfrac{1}{3}\right] \).
• Step 4 (first four nonzero terms): Expand to get \( \ln(1+3x)=3x-\dfrac{(3x)^2}{2}+\dfrac{(3x)^3}{3}-\dfrac{(3x)^4}{4}+\cdots \). Simplify coefficients: \( 3x-\dfrac{9}{2}x^2+9x^3-\dfrac{81}{4}x^4+\cdots \). State the final result as the series with its interval \( \left(-\tfrac{1}{3},\,\tfrac{1}{3}\right] \) and radius \( R=\tfrac{1}{3} \).

Problem 2 (Taylor Approximation with a Lagrange Error Bound)

• Question: Approximate \( e^{0.5} \) using the third-degree and then the sixth-degree Maclaurin polynomials for \( e^x \). Find a degree \( n \) so that the error is less than \( 10^{-4} \) and justify it using the Lagrange remainder.
• Step 1 (setup): The Maclaurin series for \( e^x \) is \( \displaystyle \sum_{k=0}^{\infty}\frac{x^k}{k!} \). The degree-\( n \) Maclaurin polynomial is \( P_n(x)=\sum_{k=0}^{n}\frac{x^k}{k!} \). The Lagrange remainder satisfies \( |R_n(x)|=\left|\dfrac{e^{c}}{(n+1)!}x^{n+1}\right| \) for some \( c \) between \( 0 \) and \( x \).
• Step 2 (degree 3 approximation): At \( x=0.5 \), \( P_3(0.5)=1+0.5+\dfrac{0.5^2}{2!}+\dfrac{0.5^3}{3!}=1+0.5+0.125+0.020833\ldots=1.645833\ldots \). The error bound is \( |R_3(0.5)|\le\dfrac{e^{0.5}}{4!}(0.5)^4 \). Since \( e^{0.5}\approx1.6487 \), \( 4!=24 \), and \( 0.5^4=0.0625 \), the bound is \( \le \dfrac{1.6487\cdot0.0625}{24}\approx0.00429 \), which is larger than \( 10^{-4} \).
• Step 3 (find \( n \) for error \( <10^{-4} \)): Use \( |R_n(0.5)|\le\dfrac{e^{0.5}}{(n+1)!}(0.5)^{n+1} \). Test \( n=5 \): \( \dfrac{e^{0.5}\cdot0.5^{6}}{6!}=\dfrac{1.6487\cdot1/64}{720}\approx4.0\times10^{-5} \), which is \( <10^{-4} \). Therefore \( n=5 \) (degree 5) suffices; using \( n=6 \) is also valid and even tighter.
• Step 4 (degree 6 approximation (tighter)): \( P_6(0.5)=\sum_{k=0}^{6}\dfrac{0.5^k}{k!}=1+0.5+0.125+0.020833+\dfrac{0.5^4}{24}+\dfrac{0.5^5}{120}+\dfrac{0.5^6}{720} \). Numerically this is \( \approx1.648721 \) to six decimals. The remainder satisfies \( |R_6(0.5)|\le\dfrac{e^{0.5}\cdot0.5^{7}}{7!}\approx2.6\times10^{-6} \), guaranteeing accuracy better than \( 10^{-4} \).