Difficulty: Intermediate to Advanced | Prerequisites: Part 1 study notes (Sections 10.1 through 10.4), geometric series, p-series, comparison tests.
This is Part 2 of the Chapter 10 material. It covers the remaining convergence tests (Ratio, Root, Alternating Series), the distinction between absolute and conditional convergence, and then shifts to power series and Taylor/Maclaurin series. Power series are where all the earlier convergence work pays off: you determine the interval on which a function can be represented as an infinite polynomial, and then use that representation to integrate, differentiate, and approximate.
The Ratio and Root Tests handle series with factorials, exponentials, or nth powers. The Alternating Series Test handles series whose terms switch sign. Absolute convergence is stronger than conditional convergence. Power series converge on an interval centred at a, whose radius R is found via the Ratio or Root Test; endpoints must be checked separately. Taylor and Maclaurin series express familiar functions (e^x, sin x, cos x, ln(1–x), arctan x) as infinite polynomials, which lets you approximate values and evaluate integrals that have no closed form.
Ratio Test
Compute L = lim(n→∞) |a_(n+1)/a_n|. If L < 1, the series converges absolutely. If L > 1 or L = ∞, it diverges. If L = 1, the test is inconclusive. Think of it as measuring whether terms shrink fast enough by comparing each term to the one before it.
Root Test
Compute L = lim(n→∞) |a_n|^(1/n). Same conclusion rules as the Ratio Test. In simple terms, it asks: what is the "average" base of the nth term when you take its nth root?
Alternating series
A series whose successive terms alternate in sign, typically written as Σ (−1)^(n−1) b_n or Σ (−1)^n b_n, where b_n > 0.
Alternating Series Test (Leibniz Test)
If {b_n} is decreasing and lim b_n = 0, then the alternating series converges. This test proves convergence only; it says nothing about divergence.
Absolute convergence
A series Σ a_n is absolutely convergent if Σ |a_n| converges. Absolute convergence implies convergence.
Conditional convergence
A series converges conditionally if Σ a_n converges but Σ |a_n| diverges. The alternating harmonic series Σ (−1)^(n−1)/n is the standard example.
Power series
A series of the form Σ c_n(x – a)^n, where x is a variable, a is the centre, and the c_n are coefficients. Think of it as an "infinite polynomial" centred at a.
Radius of convergence (R)
The value R such that the power series converges for |x – a| < R and diverges for |x – a| > R.
Interval of convergence (IOC)
The set of all x values for which the power series converges. It is (a – R, a + R) plus whichever endpoints (if any) give convergence when tested individually.
Taylor series
The power series representation of f(x) centred at a: Σ f^(n)(a)/n! · (x – a)^n.
Maclaurin series
A Taylor series centred at a = 0: Σ f^(n)(0)/n! · x^n. In simple terms, it is the special case where you expand about the origin.
Compute L = lim(n→∞) |a_(n+1) / a_n|.
L < 1: the series is absolutely convergent.
L > 1 or L = ∞: the series is divergent.
L = 1: inconclusive (could be absolutely convergent, conditionally convergent, or divergent).
The Ratio Test works best on series involving factorials, exponentials, or products of both.
Compute L = lim(n→∞) |a_n|^(1/n).
Same decision rules as the Ratio Test. The Root Test works best when the entire general term is raised to the nth power.
Example 1: Σ n³/4^n.
L = lim (n+1)³/4^(n+1) · 4^n/n³ = lim [(n+1)/n]³ · (1/4) = 1/4 < 1. Converges.
Example 2: Σ n!/4^n.
L = lim (n+1)!/4^(n+1) · 4^n/n! = lim (n+1)/4 = ∞ > 1. Diverges.
Example 3: Σ (−3)^n / n³.
L = lim |(−3)^(n+1)/(n+1)³| · |n³/(−3)^n| = 3 · lim [n/(n+1)]³ = 3 > 1. Diverges.
Factorials (n!): use the Ratio Test. The (n+1)!/n! cancellation is clean.
nth powers like (something)^n: the Root Test often simplifies nicely.
Polynomials or rational functions of n only (no factorial, no exponential): the Ratio and Root Tests both give L = 1 (inconclusive). Use the Comparison Tests or the Integral Test instead.
An alternating series has successive terms of opposite sign. Common forms for the alternating factor: (−1)^(n−1), (−1)^n, (−1)^(n+1), cos(nπ), sin((2n−1)π/2).
If the alternating series Σ (−1)^(n−1) b_n (with b_n > 0) satisfies both:
lim(n→∞) b_n = 0, and
{b_n} is eventually decreasing,
then the series converges.
This test proves convergence only. If lim b_n ≠ 0, the Test for Divergence (not the AST) tells you the series diverges.
Σ (−1)^(n−1)/n = 1 – 1/2 + 1/3 – 1/4 + ... Here b_n = 1/n. Since 1/n → 0 and {1/n} is decreasing, the series converges by the AST. Its sum is ln 2.
Σ (−1)^(n+1) n²/(n²+5). Here lim n²/(n²+5) = 1 ≠ 0, so the AST does not apply. In fact, lim a_n does not exist (the terms oscillate between values near +1 and −1), so the series diverges by the Test for Divergence.
Absolutely convergent: Σ a_n converges and Σ |a_n| also converges. Absolute convergence implies convergence.
Conditionally convergent: Σ a_n converges but Σ |a_n| diverges.
First test Σ |a_n| using any test for positive series (p-series, Ratio, Root, Comparison, Integral).
If Σ |a_n| converges: the original series is absolutely convergent. Done.
If Σ |a_n| diverges: go back to Σ a_n and try the Alternating Series Test.
If the AST gives convergence: the series is conditionally convergent.
If the Test for Divergence gives divergence: the series is divergent.
Absolutely convergent series can be rearranged in any order without changing their sum. Conditionally convergent series cannot: the Riemann Rearrangement Theorem says that for any real number r, there exists a rearrangement of a conditionally convergent series that sums to r. The alternating harmonic series is the classic illustration: its standard ordering sums to ln 2, but rearranging the terms can produce any real number you like.
Σ cos(nπ)/n^(3/2) = Σ (−1)^n / n^(3/2). Test Σ |a_n| = Σ 1/n^(3/2), a convergent p-series (p = 3/2 > 1). So the series is absolutely convergent.
A power series centred at a is Σ c_n(x – a)^n = c₀ + c₁(x–a) + c₂(x–a)² + ...
When a = 0, this simplifies to Σ c_n x^n. The sum is a function f(x) whose domain is the set of x values for which the series converges.
Use the Ratio Test (or Root Test) on the general term. Set the resulting limit < 1 and solve for x:
You will get |x – a| < R. This gives the radius of convergence R.
The open interval (a – R, a + R) is guaranteed to converge.
Plug in each endpoint x = a – R and x = a + R separately. Each endpoint produces a series of constants that you test for convergence using any test (p-series, AST, Test for Divergence, etc.).
The interval of convergence uses square brackets for endpoints that converge and parentheses for endpoints that diverge.
Special cases: R = 0 means the series converges only at x = a. R = ∞ means the series converges for all x.
Ratio Test: L = lim |a_(n+1)/a_n| = lim [(n+1)²/2^(n+1)] · [2^n/(n²)] · |x| = |x|/2 · lim [(n+1)/n]² = |x|/2.
For convergence: |x|/2 < 1, so |x| < 2. Radius R = 2.
Check x = 2: Σ (−1)^n n² (2)^n / 2^n = Σ (−1)^n n². Since lim (−1)^n n² does not exist, this diverges.
Check x = −2: Σ (−1)^n n² (−2)^n / 2^n = Σ n². Since lim n² = ∞, this diverges.
Interval of convergence: (−2, 2).
The geometric series gives the starting point: 1/(1–x) = Σ x^n = 1 + x + x² + x³ + ..., valid for |x| < 1.
By substitution, differentiation, and integration, you can build representations for many functions. For example:
Replace x with –x: 1/(1+x) = Σ (−1)^n x^n, |x| < 1.
Differentiate: 1/(1–x)² = Σ n x^(n−1), |x| < 1.
Integrate: –ln(1–x) = Σ x^(n+1)/(n+1), |x| < 1.
Replace x with –x²: 1/(1+x²) then integrate to get arctan x.
Suppose f(x) has a power series representation f(x) = Σ c_n(x–a)^n for |x–a| < R. By repeatedly differentiating and evaluating at x = a, the coefficients turn out to be:
c_n = f^(n)(a) / n!
This gives the Taylor series of f centred at a:
f(x) = f(a) + f′(a)(x–a) + f″(a)(x–a)²/2! + f″′(a)(x–a)³/3! + ...
When a = 0, the result is called a Maclaurin series.
All derivatives of e^x are e^x, so f^(n)(0) = 1 for every n.
e^x = Σ x^n/n! = 1 + x + x²/2! + x³/3! + x⁴/4! + ... with R = ∞.
The derivatives cycle through sin, cos, –sin, –cos. Evaluating at 0 gives 0, 1, 0, −1, 0, 1, ... Only odd powers survive.
sin x = Σ (−1)^n x^(2n+1) / (2n+1)! = x – x³/3! + x⁵/5! – x⁷/7! + ... with R = ∞.
Similarly, only even powers survive.
cos x = Σ (−1)^n x^(2n) / (2n)! = 1 – x²/2! + x⁴/4! – x⁶/6! + ... with R = ∞.
1/(1–x) = Σ x^n = 1 + x + x² + x³ + ..., R = 1
1/(1–x)² = Σ n x^(n−1) = 1 + 2x + 3x² + 4x³ + ..., R = 1
ln(1–x) = –Σ x^(n+1)/(n+1) = –x – x²/2 – x³/3 – ..., R = 1
arctan x = Σ (−1)^n x^(2n+1)/(2n+1) = x – x³/3 + x⁵/5 – ..., R = 1
e^x = Σ x^n/n! = 1 + x + x²/2! + x³/3! + ..., R = ∞
sin x = Σ (−1)^n x^(2n+1)/(2n+1)! = x – x³/3! + x⁵/5! – ..., R = ∞
cos x = Σ (−1)^n x^(2n)/(2n)! = 1 – x²/2! + x⁴/4! – ..., R = ∞
The nth-degree Taylor polynomial T_n(x) is the partial sum of the Taylor series truncated at the x^n term. It provides a polynomial approximation of f(x) near x = a. Higher degree means better accuracy over a wider interval.
Example: the third-degree Taylor polynomial for f(x) = √x centred at x = 1 is T₃(x) = 1 + (1/2)(x−1) – (1/8)(x−1)² + (1/16)(x−1)³. Using this: T₃(1.5) ≈ 1.2266, while the true value is √1.5 ≈ 1.2247.
When no closed-form antiderivative exists, substitute the Maclaurin series and integrate term by term.
Example: ∫ x³ e^(–x³) dx. Replace e^(–x³) with its series: 1 – x³ + x⁶/2! – x⁹/3! + ... Multiply by x³: x³ – x⁶ + x⁹/2 – x¹²/6 + ... Integrate term by term: x⁴/4 – x⁷/7 + x¹⁰/20 – x¹³/78 + ... + C.
Ratio Test
L = lim(n→∞) |a_(n+1) / a_n|. Converges absolutely if L < 1; diverges if L > 1.
Root Test
L = lim(n→∞) |a_n|^(1/n). Same rules as the Ratio Test.
Alternating Series Test conditions
Σ (−1)^(n−1) b_n converges if (i) lim b_n = 0 and (ii) {b_n} is eventually decreasing.
Taylor series centred at a
f(x) = Σ f^(n)(a)/n! · (x–a)^n
Maclaurin series (a = 0)
f(x) = Σ f^(n)(0)/n! · x^n
Radius of convergence via Ratio Test
Set lim |a_(n+1)/a_n| < 1, solve for |x–a| < R.
Students often think the Ratio Test works on every series. It does not: for purely rational functions of n (no factorial or exponential), it gives L = 1 (inconclusive). Use Comparison or Integral Tests instead.
The Alternating Series Test proves convergence only. If it fails (because lim b_n ≠ 0), the correct next step is the Test for Divergence, not a conclusion that the series diverges "by AST."
Students confuse absolute convergence with convergence. Absolute convergence is the stronger condition. A series can converge without converging absolutely (that is conditional convergence).
When finding the interval of convergence, students sometimes forget to test the endpoints separately. The Ratio Test tells you nothing at |x–a| = R; you must substitute and use a different test.
Students sometimes think a Taylor polynomial is the Taylor series. The polynomial T_n(x) is an approximation (finitely many terms); the series is the full infinite sum.
⚠️ "Determine whether the series is absolutely convergent, conditionally convergent, or divergent" is the most common phrasing on exams for this material. Follow the classification flowchart from Section 10.6.
⚠️ Expect at least one problem asking you to find the radius and interval of convergence of a power series. Always test the endpoints.
⚠️ You will likely be asked to write the Maclaurin series for e^x, sin x, or cos x from scratch using the derivative table. Memorise the patterns so you can move quickly.
⚠️ The list of important Maclaurin series is heavily tested. Know the formula and the radius of convergence for each.
⚠️ Using a power series to integrate a function (like ∫ x³ e^(–x³) dx) is a common free-response problem. Substitute the series, multiply, then integrate term by term.
True or False: The Ratio Test is inconclusive for the series Σ 1/n². True (L = 1 for any p-series).
Fill in the blank: A series is conditionally convergent if Σ a_n converges but Σ ____ diverges. |a_n|.
True or False: The Maclaurin series for e^x converges for all real numbers. True (R = ∞).
Fill in the blank: The radius of convergence of Σ x^n is R = ____. 1.
True or False: If the Alternating Series Test fails, the series must diverge. False (the AST only proves convergence; its failure means the test is inapplicable, not that the series diverges, though in practice if lim b_n ≠ 0 you then apply the Test for Divergence).
Q: Determine whether Σ (−1)^(n−1) / √n is absolutely convergent, conditionally convergent, or divergent.
A: Test Σ |a_n| = Σ 1/√n = Σ 1/n^(1/2), a divergent p-series (p = 1/2). So it is not absolutely convergent. Now test Σ (−1)^(n−1)/√n with the AST: b_n = 1/√n is decreasing and lim 1/√n = 0. The series converges by the AST. Conclusion: conditionally convergent.
Q: Find the Maclaurin series for f(x) = e^(−2x) and state its radius of convergence.
A: Start with e^x = Σ x^n/n!. Replace x with −2x: e^(−2x) = Σ (−2x)^n/n! = Σ (−1)^n 2^n x^n / n!. The radius of convergence is R = ∞ (same as e^x).
Q: Find the radius and interval of convergence of Σ (x+4)^n / (n · 3^n).
A: Ratio Test: L = |x+4|/3. For convergence: |x+4| < 3, so R = 3 and the open interval is (∓77/3, ∑1/3), i.e. (∓7, −1). Check x = −1: Σ 3^n/(n · 3^n) = Σ 1/n, the harmonic series (diverges). Check x = −7: Σ (−1)^n/n, which converges by the AST. IOC: [−7, −1).
Q: Use a known Maclaurin series to find the power series for arctan x. What is R?
A: Start with 1/(1+x²) = Σ (−1)^n x^(2n). Integrate term by term: arctan x = Σ (−1)^n x^(2n+1)/(2n+1). R = 1.
Q: Use the Ratio Test on Σ (2n)! / (n!)^2. Does it converge or diverge?
A: a_(n+1)/a_n = (2n+2)!/(((n+1)!)²) · (n!)²/(2n)! = (2n+2)(2n+1)/((n+1)²) = (2(2n+1))/(n+1). As n → ∞, this approaches 4. Since 4 > 1, the series diverges.
The Ratio and Root Tests build directly on the geometric series convergence condition |r| < 1 from Section 10.2. The Alternating Series Test uses the monotonicity and limit ideas from Section 10.1 on sequences. Power series bring together every convergence test from the course, since checking endpoints may require the p-series test, the AST, or the Test for Divergence. Taylor and Maclaurin series connect back to derivatives (Chapter 3) and integration techniques (Chapters 7 and 8), and are foundational for Calculus 2 topics like differential equations solved via series and Fourier series.
Ratio Test, Root Test, Alternating Series Test, Leibniz Test, absolute convergence, conditional convergence, Riemann Rearrangement Theorem, power series, radius of convergence, interval of convergence, ROC, IOC, Taylor series, Maclaurin series, Taylor polynomial, e^x series, sin x series, cos x series, ln(1–x) series, arctan x series, 1/(1–x) series, term-by-term integration, term-by-term differentiation, Math 104, Penn State, Calculus 1, Chapter 10