An infinite series $\displaystyle\sum_{n=1}^{\infty} a_n$ is the limit of partial sums $S_N = \sum_{n=1}^N a_n$. If $\lim_{N\to\infty} S_N$ exists and is finite, the series converges; otherwise it diverges.
Geometric Series: $\displaystyle\sum_{n=0}^{\infty} ar^n = \dfrac{a}{1-r}$ converges iff $|r| < 1$.
$p$-Series: $\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^p}$ converges iff $p > 1$. (Harmonic series $p=1$ diverges.)
Divergence Test (necessary condition): If $\displaystyle\sum a_n$ converges, then $\lim_{n\to\infty} a_n = 0$. Contrapositive: if $\lim a_n \neq 0$, the series definitely diverges.
Find the sum of $\displaystyle\sum_{n=0}^{\infty} \frac{3}{2^n} = 3\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^n$.
$|r| = 1/2 < 1$ ✓. Sum $= \dfrac{3}{1-1/2} = \dfrac{3}{1/2} = 6$.
$L = \lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|$
$L < 1$: converges absolutely
$L > 1$: diverges
$L = 1$: inconclusive
Best for: factorials, exponentials
If $f(x)$ is positive, decreasing, continuous: $\sum a_n$ and $\int_1^\infty f(x)\,dx$ both converge or both diverge.
Best for: $p$-series type terms
If $0 \leq a_n \leq b_n$:
$\sum b_n$ converges → $\sum a_n$ converges
$\sum a_n$ diverges → $\sum b_n$ diverges
Best for: rational expressions
$\sum(-1)^n b_n$ converges if $b_n$ is decreasing and $b_n\to 0$.
Error bound: $|S - S_N| \leq b_{N+1}$
Best for: series with $(-1)^n$
$\dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)!}{3^{n+1}} \cdot \dfrac{3^n}{n!} = \dfrac{n+1}{3} \to \infty$ as $n\to\infty$.
Since $L = \infty > 1$: the series diverges.
$b_n = 1/n$ is decreasing and $b_n \to 0$. Series converges by AST.
This is the alternating harmonic series, which sums to $\ln 2$.
Error estimate: Using $S_4 = 1 - 1/2 + 1/3 - 1/4 = 7/12$: error $\leq b_5 = 1/5 = 0.2$. (Actual $\ln 2 \approx 0.693$; $S_4 = 0.583$.)
Determine convergence of $\displaystyle\sum_{n=1}^{\infty} \dfrac{2^n}{n!}$ using the Ratio Test.
AP Exam Tip: The Ratio Test is the workhorse for BC series problems — use it first when you see factorials or exponentials. The Divergence Test should always be checked first (it's fastest). The AP exam favors series that require exactly one clean test; if the ratio test gives $L=1$, try the comparison or integral test.
Partial sums $S_N$ of the alternating harmonic series converge to $\ln 2 \approx 0.693$. Notice how successive sums alternately overshoot and undershoot the limit.
Figure 11.1 — Partial Sums of $\sum \dfrac{(-1)^{n+1}}{n}$ Converging to $\ln 2$
A power series centered at $a$ has the form $\displaystyle\sum_{n=0}^{\infty} c_n(x-a)^n$. It converges for $x$ in some interval of convergence centered at $a$ with radius of convergence $R$.
Find the interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \dfrac{(x-2)^n}{n \cdot 3^n}$.
Ratio Test: $\left|\dfrac{(x-2)^{n+1}}{(n+1)3^{n+1}} \cdot \dfrac{n\cdot3^n}{(x-2)^n}\right| = \dfrac{n}{n+1} \cdot \dfrac{|x-2|}{3} \to \dfrac{|x-2|}{3}$
Converges when $\dfrac{|x-2|}{3} < 1$, i.e., $|x-2| < 3$, so $-1 < x < 5$.
Endpoint $x=5$: $\sum \dfrac{3^n}{n\cdot3^n} = \sum\dfrac{1}{n}$ — diverges (harmonic).
Endpoint $x=-1$: $\sum \dfrac{(-3)^n}{n\cdot3^n} = \sum\dfrac{(-1)^n}{n}$ — converges (alternating harmonic).
Interval of convergence: $[-1, 5)$, radius $R = 3$.
A Taylor series expresses a function as an infinite polynomial, making it possible to compute values, take derivatives, and integrate even complex functions.
$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots$$
A Maclaurin series is a Taylor series at $a = 0$.
You must memorize these five Maclaurin series for the AP BC exam:
| Function | Maclaurin Series | Radius of Convergence |
|---|---|---|
| $e^x$ | $\displaystyle\sum_{n=0}^{\infty}\dfrac{x^n}{n!} = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots$ | $R = \infty$ |
| $\sin x$ | $\displaystyle\sum_{n=0}^{\infty}\dfrac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \dfrac{x^3}{6} + \dfrac{x^5}{120} - \cdots$ | $R = \infty$ |
| $\cos x$ | $\displaystyle\sum_{n=0}^{\infty}\dfrac{(-1)^n x^{2n}}{(2n)!} = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} - \cdots$ | $R = \infty$ |
| $\dfrac{1}{1-x}$ | $\displaystyle\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots$ | $R = 1$ |
| $\ln(1+x)$ | $\displaystyle\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}x^n}{n} = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \cdots$ | $R = 1$ |
Find the first four terms of the Maclaurin series for $f(x) = e^{-x^2}$.
Replace $x$ with $-x^2$ in the series for $e^x$:
$$e^{-x^2} = 1 + (-x^2) + \frac{(-x^2)^2}{2!} + \frac{(-x^2)^3}{3!} + \cdots = 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6} + \cdots$$
This is far easier than differentiating $e^{-x^2}$ repeatedly!
Write the first three non-zero terms of the Maclaurin series for $\sin(x^2)$.
Use the degree-3 Taylor polynomial at $a=0$ for $e^x$ to approximate $e^{0.2}$, and bound the error.
$P_3(x) = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{6}$
$P_3(0.2) = 1 + 0.2 + 0.02 + \dfrac{0.008}{6} = 1.2213\overline{3}$
Lagrange Error Bound: $|R_3(x)| \leq \dfrac{M}{(3+1)!}|x|^4$ where $M$ = max of $|f^{(4)}|$ on $[0, 0.2]$.
$f^{(4)}(x) = e^x \leq e^{0.2} < 2$, so $|R_3| \leq \dfrac{2}{24}(0.2)^4 = \dfrac{2}{24}\cdot 0.0016 \approx 0.000133$.
Actual $e^{0.2} \approx 1.22140$ — error $\approx 0.00007 <$ bound ✓.
AP Exam Tip: BC FRQ series problems have a predictable structure: (a) write the first few terms, (b) find the general term, (c) determine interval of convergence, (d) use the series to approximate a value. The Lagrange Error Bound is frequently tested — remember that it always overestimates the actual error.
Taylor polynomials for $e^x$ converging to the true function. Add more terms (increase $N$) to see the polynomial match $e^x$ over a wider interval.
Figure 11.2 — Taylor Polynomials for $e^x$ of Increasing Degree
Taylor polynomial for $\sin x$ — notice how accuracy improves near $x=0$ and degrades far away. Slider controls the degree.
Figure 11.3 — Taylor Polynomials for $\sin x$ at $a = 0$
Determine whether $\displaystyle\sum_{n=1}^{\infty}\dfrac{n^2}{n^3+1}$ converges or diverges.
Use the Ratio Test to determine convergence of $\displaystyle\sum_{n=0}^{\infty}\dfrac{3^n}{(2n)!}$.
Find the interval of convergence of $\displaystyle\sum_{n=0}^{\infty}\dfrac{x^n}{n+1}$.
Write the Maclaurin series for $\dfrac{1}{1+x^2}$ and use it to derive a series for $\arctan x$.
Estimate $\displaystyle\int_0^1 e^{-x^2}\,dx$ using the first three terms of the Maclaurin series for $e^{-x^2}$.
Find the sum of the series $\displaystyle\sum_{n=0}^{\infty}\dfrac{(-1)^n\pi^{2n+1}}{4^{2n+1}(2n+1)!}$.
Use the Lagrange Error Bound to determine how many terms of the Maclaurin series for $\cos x$ are needed to approximate $\cos(0.5)$ with error less than $0.0001$.
(AP Style) The function $f$ has a Maclaurin series with $f^{(n)}(0) = \dfrac{(-1)^n \cdot n!}{2^n}$. (a) Write the first four terms. (b) Find a closed form for $f(x)$. (c) Find the IOC.
If $\lim_{n\to\infty} a_n \neq 0$, then $\sum a_n$ diverges. (Note: $a_n \to 0$ does NOT guarantee convergence.)
Let $L = \lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|$. Converges if $L<1$, diverges if $L>1$, inconclusive if $L=1$.
$\sum(-1)^n b_n$ converges if $b_n$ is decreasing and $b_n \to 0$. Error bound: $|S - S_N| \leq b_{N+1}$.
$\sum \dfrac{1}{n^p}$ converges if $p > 1$, diverges if $p \leq 1$. Harmonic series ($p=1$) diverges.
$\displaystyle\sum_{n=0}^\infty ar^n = \dfrac{a}{1-r}$ for $|r|<1$. Diverges if $|r|\geq 1$.
$f(x) = \displaystyle\sum_{n=0}^\infty \dfrac{f^{(n)}(a)}{n!}(x-a)^n$ — centered at $x = a$.
$e^x = \sum\dfrac{x^n}{n!}$ $\sin x = \sum\dfrac{(-1)^n x^{2n+1}}{(2n+1)!}$ $\cos x = \sum\dfrac{(-1)^n x^{2n}}{(2n)!}$
Found via Ratio Test on power series $\sum c_n(x-a)^n$: $R = \lim_{n\to\infty}\left|\dfrac{c_n}{c_{n+1}}\right|$.