Core Concepts
Concept 01
Ratio Test
Compute \(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=L\). Converges if \(L<1\), diverges if \(L>1\).
RATIO → RADIUS
Concept 02
Taylor Series Term
The \(n\)-th term is \(\dfrac{f^{(n)}(a)}{n!}(x-a)^n\). Center \(a=4\) here.
CENTER → SHIFT
Concept 03
Differentiating a Series
Differentiate term-by-term. The radius of convergence stays the same; endpoints may change.
TERM-BY-TERM
Concept 04
Geometric Series Sum
\(\sum_{n=0}^{\infty} ar^n = \dfrac{a}{1-r}\) when \(|r|<1\).
a / (1 − r)
Concept 05
Endpoint Behavior
Ratio test is inconclusive at endpoints (\(L=1\)). Test separately with known tests (p-series, AST, etc.).
CHECK ENDS
Level 1 · Foundations
What does the Ratio Test check?
Select the most precise description of the Ratio Test for a series \(\sum a_n\).
KEY: RATIO → RADIUS — the ratio test gives you the radius directly once you solve \(L < 1\).
Variation
If \(L = 1\) in the Ratio Test, what can you conclude? Why does this matter at the endpoints of an interval of convergence?
Identify the general term of the given Taylor series.
The series is \(\displaystyle\sum_{n=1}^{\infty} \frac{(x-4)^{n+1}}{(n+1)3^n}\). What is the general term \(a_n\)?
KEY: CENTER → SHIFT — whenever you see \((x-4)\), the series is centered at \(a = 4\). Identify numerator and denominator separately.
Variation
Write out \(a_{n+1}\) explicitly for this series. You'll need it for the Ratio Test in Q3.
Form the ratio \(\left|\dfrac{a_{n+1}}{a_n}\right|\).
For \(a_n = \dfrac{(x-4)^{n+1}}{(n+1)3^n}\), which expression correctly simplifies \(\left|\dfrac{a_{n+1}}{a_n}\right|\)?
KEY: FLIP & MULTIPLY — dividing by a fraction means multiplying by its reciprocal. Cancel like terms top and bottom.
Variation
What is \(\lim_{n\to\infty}\dfrac{n+1}{n+2}\)? Why does this limit always appear and what does it simplify to?
Take the limit and find the Radius of Convergence.
Given \(\lim_{n\to\infty}\dfrac{n+1}{n+2}\cdot\dfrac{|x-4|}{3} = L\), what is the radius of convergence \(R\)?
KEY: SET L < 1 → SOLVE → RADIUS — after taking the limit, set the expression less than 1 and solve for \(|x-4|\).
Variation
If the general term had been \(\dfrac{(x-4)^{n+1}}{(n+1)\cdot 5^n}\) instead, what would the radius of convergence be?
Write the interval of convergence in inequality form (before checking endpoints).
The series is centered at \(x = 4\) with radius \(R = 3\). What is the open interval (ignoring endpoints)?
KEY: CENTER ± RADIUS — interval = \((a - R,\ a + R)\). Always write this before testing endpoints.
Variation
A Taylor series is centered at \(x = -2\) and has radius of convergence \(R = 5\). Write the open interval of convergence.
Level 2 · Intermediate
Test the endpoint \(x = 1\).
Substitute \(x = 1\) into the series \(\displaystyle\sum_{n=1}^{\infty}\frac{(x-4)^{n+1}}{(n+1)3^n}\). Which series do you get, and does it converge?
KEY: CHECK ENDS — at \(x=1\): \((1-4)^{n+1} = (-3)^{n+1}\). Watch signs carefully. Use AST or p-series.
Variation
Now test the endpoint \(x = 7\). What series do you get? Does it converge or diverge?
State the complete interval of convergence including endpoints.
Given that \(x = 1\) converges and \(x = 7\) produces a divergent harmonic-type series, what is the final interval of convergence?
KEY: BRACKETS = INCLUDED — use \([\) for endpoints that converge, \((\) for those that diverge.
Variation
If both endpoints converged, what would the interval of convergence look like? If both diverged?
Differentiate the series term-by-term to find the Taylor series for \(f'\).
Given \(f(x) = \displaystyle\sum_{n=1}^{\infty}\frac{(x-4)^{n+1}}{(n+1)3^n}\), what is the first term of \(f'(x)\)?
KEY: TERM-BY-TERM — differentiate each term: \(\dfrac{d}{dx}(x-4)^{n+1} = (n+1)(x-4)^n\). The \((n+1)\) cancels!
Variation
Compute the second and third terms of \(f'(x)\) by differentiating the \(n=2\) and \(n=3\) terms of \(f\).
Find the general term of \(f'(x)\).
After differentiating term-by-term, which expression gives the correct general term of \(f'(x)\)?
KEY: (n+1) CANCELS — \(\dfrac{d}{dx}\left[\dfrac{(x-4)^{n+1}}{(n+1)3^n}\right] = \dfrac{(x-4)^n}{3^n} = \left(\dfrac{x-4}{3}\right)^n\). Beautiful cancellation!
Variation
The general term of \(f'(x)\) is \(\left(\dfrac{x-4}{3}\right)^n\). What type of series is \(\sum_{n=1}^{\infty}\left(\dfrac{x-4}{3}\right)^n\)? What is the common ratio \(r\)?
Sum the geometric series for \(f'(x)\).
The series \(f'(x) = \displaystyle\sum_{n=1}^{\infty}\left(\dfrac{x-4}{3}\right)^n\) is geometric with first term \(a = \dfrac{x-4}{3}\) and ratio \(r = \dfrac{x-4}{3}\). What is its sum?
KEY: a / (1 − r) — geometric sum starts at \(n=1\): \(\sum_{n=1}^{\infty}ar^{n-1} = \dfrac{a}{1-r}\). Or use \(\dfrac{r}{1-r}\) when first term is \(r\).
Variation
Verify the algebra: simplify \(\dfrac{(x-4)/3}{1-(x-4)/3}\) step-by-step to confirm you get \(\dfrac{x-4}{7-x}\).
Level 3 · AP Difficulty
Does the Taylor series for \(f'\) converge to \(f'(x) = \dfrac{x-4}{7-x}\) at \(x = 8\)?
The radius of convergence for both \(f\) and \(f'\) is \(R = 3\), centered at \(x = 4\). The interval of convergence of the series for \(f'\) is \((1, 7)\). At \(x = 8\), which statement is correct?
KEY: OUTSIDE IOC → NO CONVERGENCE — if \(x\) is outside the interval of convergence, the series diverges and cannot equal \(f'(x)\).
Variation
What if the problem asked about \(x = 4\)? Does the series converge to \(f'(4)\) there? What is \(f'(4)\)?
What are the first three nonzero terms of the Taylor series for \(f'\) about \(x = 4\)?
The series is \(f'(x) = \displaystyle\sum_{n=1}^{\infty}\left(\dfrac{x-4}{3}\right)^n\). Write out the first three terms explicitly.
KEY: PLUG IN n = 1, 2, 3 — substitute \(n=1, 2, 3\) into the general term and simplify each.
Variation
Identify the common ratio \(r\) of this geometric series. In the formula \(\frac{a}{1-r}\), what is \(a\) and what is \(r\)?
Show that the geometric series for \(f'\) sums to \(\dfrac{x-4}{7-x}\) algebraically.
Which step in the simplification of \(\dfrac{(x-4)/3}{1-(x-4)/3}\) is incorrect?
KEY: COMMON DENOMINATOR — \(1 - \dfrac{x-4}{3} = \dfrac{3-(x-4)}{3} = \dfrac{7-x}{3}\). Then divide.
Variation
Verify by plugging \(x = 5\) into both \(\sum_{n=1}^{\infty}((5-4)/3)^n\) (compute a few partial sums) and \(\dfrac{5-4}{7-5} = \dfrac{1}{2}\). Do they agree?
A Taylor series for \(g\) about \(x = 2\) has general term \(\dfrac{(-1)^n (x-2)^n}{n \cdot 4^n}\). Find the interval of convergence.
Apply the Ratio Test, then check both endpoints carefully.
KEY: ALTERNATING AT ONE END — the \((-1)^n\) factor often makes one endpoint an alternating series (AST) and the other a regular p-series or harmonic.
Variation
How would the interval of convergence change if the general term were \(\dfrac{(-1)^n(x-2)^n}{n^2 \cdot 4^n}\)? (Note: \(n^2\) instead of \(n\).)
The Taylor series for \(h(x) = \ln(x)\) centered at \(x = 1\) is \(\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}(x-1)^n}{n}\). What is the interval of convergence?
Use the Ratio Test to find \(R\), then test endpoints \(x = 0\) and \(x = 2\).
KEY: ln(x) SERIES — classic AP series. At \(x=2\): alternating harmonic (converges). At \(x=0\): harmonic with signs cancelled (diverges).
Variation
The series for \(\ln(1+x)\) centered at \(x=0\) is \(\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}x^n}{n}\). What is its interval of convergence?
Level 4 · AP Free Response
Part A (Ratio Test + Full Justification): What is the correct justification sequence for finding the interval of convergence of the given series?
For \(\displaystyle\sum_{n=1}^{\infty}\frac{(x-4)^{n+1}}{(n+1)3^n}\), an AP reader expects a complete, logical argument. Which sequence of steps is fully correct and complete?
KEY: RATIO → LIMIT → SET < 1 → SOLVE → TEST ENDS — this 5-step flow earns full AP credit every time.
AP Free Response Tip
In AP scoring, "justification" means showing the limit, setting it less than 1, solving the inequality, and testing both endpoints with named tests. Missing any step typically costs a point.
Part B: Which of the following correctly states the first three nonzero terms and general term of \(f'\)?
The series for \(f'\) is \(\displaystyle\sum_{n=1}^{\infty}\left(\dfrac{x-4}{3}\right)^n\). Write in the AP free-response format.
KEY: WRITE ALL PARTS — AP Part B typically wants: first term + second term + third term + "..." + general term. Don't skip any.
AP Free Response Tip
AP readers look for the "+···+" between the third term and the general term. Writing just three terms without a general term loses a point. Always include the general term explicitly.
Part C: Which proof correctly shows \(f'(x) = \dfrac{x-4}{7-x}\) from the geometric series?
For full AP credit, the argument must identify the geometric series, state the formula used, and complete the algebra.
KEY: NAME THE TYPE — say "geometric series with first term \(a\) and ratio \(r\)" before applying \(a/(1-r)\). AP rewards naming.
AP Free Response Tip
Always state "converges for \(|r| < 1\)" when applying the geometric series formula. This connects to the interval of convergence and earns justification credit.
Part D (Synthesis): The series for \(f'\) has the same radius of convergence as \(f\). Why is this always true?
Choose the most precise and mathematically correct explanation.
KEY: SAME RADIUS, MAYBE DIFFERENT ENDPOINTS — term-by-term differentiation preserves radius. But endpoints can change.
Variation
Give an example where the interval of convergence of \(f'\) differs from that of \(f\) (i.e., they have the same radius but different endpoint inclusion).
Full synthesis: A new function \(F(x) = \displaystyle\int_4^x f(t)\,dt\). If \(f(x) = \displaystyle\sum_{n=1}^{\infty}\frac{(x-4)^{n+1}}{(n+1)3^n}\), find the general term of the Taylor series for \(F\).
Integrate term-by-term. What is the general term of \(F(x)\)?
KEY: INTEGRATE TERM-BY-TERM — \(\int_4^x (t-4)^{n+1}\,dt = \dfrac{(x-4)^{n+2}}{n+2}\). The \((n+2)\) goes in the denominator.
AP Free Response Tip
This type of problem appears often: given a series for \(f\), find a series for \(\int f\). Always integrate term-by-term from the center (lower limit = center), so the constant of integration is 0 if \(F(a) = 0\).