Home / AP Calculus BC: 10.2 Working with  Geometric Series – Exam Style questions with Answer- FRQ

AP Calculus BC: 10.2 Working with  Geometric Series – Exam Style questions with Answer- FRQ

Question

(a) Topic-10.8 Ratio Test for Convergence

(b) Topic-10.10 Alternating Series Error Bound

(c) Topic-10.14 Finding Taylor or Maclaurin Series for a Function

(d) Topic-10.2 Working with Geometric Series

The function $f$ is defined by the power series
\(
f(x) = x – \frac{x^3}{3} + \frac{x^5}{5} – \frac{x^7}{7} + \cdots + \frac{(-1)^n x^{2n+1}}{2n+1} + \cdots
\)
for all real numbers $x$ for which the series converges.
(a) Using the ratio test, find the interval of convergence of the power series for $f$. Justify your answer.
(b) Show that
\(
\left| f\left(\frac{1}{2}\right) – \frac{1}{2} \right| < \frac{1}{10}.
\)
Justify your answer.
(c) Write the first four nonzero terms and the general term for an infinite series that represents $f'(x)$.
(d) Use the result from part (c) to find the value of $f’\left(\frac{1}{6}\right)$.

▶️Answer/Explanation

\(\textbf{6(a)}\)
Using the ratio test, we want to find all $x$ such that
\(
\lim_{n \to \infty} \left| \frac{x^{2(n+1)+1}}{2(n+1)+1} \cdot \frac{2n+1}{x^{2n+1}} \right|
= \lim_{n \to \infty} \left| \frac{x^2 (2n+1)}{2n+3} \right| < 1.
\)
\(
\lim_{n \to \infty} \frac{2n+1}{2n+3} = 1, \quad \text{so } |x|^2 < 1 \implies -1 < x < 1,
\)
and the radius of convergence is $1$.

Testing the endpoints:

When $x = -1$:
\(
\sum_{n=0}^\infty \frac{(-1)^n (-1)^{2n+1}}{2n+1} = \sum_{n=0}^\infty \frac{(-1)^{n+1}}{2n+1}.
\)

When $x = 1$:
\(
\sum_{n=0}^\infty \frac{(-1)^n (1)^{2n+1}}{2n+1} = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}.
\)

Both are alternating series whose terms decrease in absolute value to $0$, so they both converge.

In other words, they are alternating series, $a_{n+1} < a_n$, and $\lim_{n \to \infty} \frac{1}{2n+1} = 0$.

So, the interval of convergence of $f$ is $-1 \leq x \leq 1$.

\(\textbf{6(b)}\)
\(
f\left(\frac{1}{2}\right) \approx \frac{1}{2}, \quad \text{so we can say that this represents } P_1\left(\frac{1}{2}\right),
\)
the first-degree Taylor polynomial for the alternating series $f(x)$ when $x = \frac{1}{2}$.

\(
\left| f(x) – \frac{1}{2} \right| = \left| f(x) – P_1\left(\frac{1}{2}\right) \right| \quad \text{is the error form for the alternating series.}
\)

Hence, the alternating series error bound is the first omitted term:
\(
\left| f(x) – \frac{1}{2} \right| = \left| f(x) – P_1\left(\frac{1}{2}\right) \right| \leq \left| \frac{\left(\frac{1}{2}\right)^3}{3} \right| = \frac{1}{24} < \frac{1}{10}.
\)

\(\textbf{(c)}\)
\(
f'(x) = 1 – \frac{3x^2}{3} + \frac{5x^4}{5} – \frac{7x^6}{7} + \cdots + \frac{(2n+1)(-1)^n x^{2n}}{2n+1} + \cdots
\)
or
\(
f'(x) = 1 – x^2 + x^4 – x^6 + \cdots + (-1)^n x^{2n} + \cdots
\)

\(\textbf{(d)}\)
\(
f’\left(\frac{1}{6}\right) \approx 1 – \left(\frac{1}{6}\right)^2 + \left(\frac{1}{6}\right)^4 – \left(\frac{1}{6}\right)^6 + \cdots
\)
This forms a geometric series where $a_1 = 1$ and $r = \left(\frac{1}{6}\right)^2$.

The sum of the series is given by:
\(
\text{Sum} = \frac{a_1}{1 – r} = \frac{1}{1 – \left(\frac{1}{6}\right)^2}.
\)

Simplify:
\(
\text{Sum} = \frac{1}{1 – \frac{1}{36}} = \frac{1}{\frac{36 – 1}{36}} = \frac{36}{35}.
\)

Thus:
\(
f’\left(\frac{1}{6}\right) = \frac{36}{35}.
\)

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