AP Calculus BC
Sequences & Series
Convergence Tests, Power Series, Taylor Series, and Error Bounds
VINAY KANTH RAO KODIPELLY
Foundations: Sequences, Series, and Partial Sums
In AP Calculus BC, a sequence is an ordered list of numbers $a_1, a_2, a_3, \dots$ and a series is the sum of the terms of a sequence:
Since we cannot physically add infinitely many numbers, we define the $n$-th partial sum as $S_n = a_1 + a_2 + \dots + a_n.$
The Core Principle
We establish as a foundational fact that a series converges to $S$ if and only if the sequence of its partial sums converges to $S$: $\lim_{n\to\infty} S_n = S$.
Critical AP Warning
The limit of the terms going to zero ($\lim_{n\to\infty} a_n = 0$) does NOT prove convergence. It is only a necessary condition. The classic counterexample is the harmonic series $\sum \frac{1}{n}$. The terms go to 0, but the series diverges.
1. The n-th Term Test for Divergence
Standard Theorem
If $\lim_{n \to \infty} a_n \neq 0$, then the series $\sum a_n$ diverges.
Caution: If $\lim_{n \to \infty} a_n = 0$, the test is inconclusive.
- Identify the expression for $a_n$.
- Take the limit as $n \to \infty$.
- If non-zero, state "Diverges by n-th term test."
- If exact 0, immediately move to another test.
Worked Examples
$\sum_{n=1}^{\infty} \frac{n}{n+1}$
Degrees equal, horizontal asymptote check.
$\lim_{n\to\infty} \frac{n}{n+1} = 1 \neq 0$$\sum_{n=1}^{\infty} \frac{3n+1}{2n-5}$
Numerator and denominator degree 1.
$\lim_{n\to\infty} \frac{3n+1}{2n-5} = 1.5 \neq 0$$\sum_{n=1}^{\infty} \cos(n\pi)$
Alternates between -1 and 1.
$\lim \cos(n\pi)$ DNE $\neq 0$$\sum_{n=0}^{\infty} \sin(n)$
Oscillates without settling.
$\lim \sin(n)$ DNE $\neq 0$$\sum_{n=1}^{\infty} \frac{n^2+1}{5n^2-3}$
Degrees equal to 2.
$\lim_{n\to\infty} \frac{n^2+1}{5n^2-3} = \frac{1}{5} \neq 0$$\sum_{n=1}^{\infty} \frac{1}{n}$
Classic harmonic series.
$\lim_{n\to\infty} \frac{1}{n} = 0$$\sum_{n=1}^{\infty} \frac{1}{n^2}$
Denominator power is higher.
$\lim_{n\to\infty} \frac{1}{n^2} = 0$$\sum_{n=2}^{\infty} \frac{\ln n}{n}$
n grows faster than ln(n). L'Hôpital.
$\lim_{n\to\infty} \frac{\ln n}{n} = 0$$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$
Denominator grows to infinity.
$\lim_{n\to\infty} \frac{1}{\sqrt{n}} = 0$$\sum_{n=1}^{\infty} \frac{1+(-1)^n}{n}$
Numerator bounded between 0 and 2.
Note that $0 \le \frac{1+(-1)^n}{n} \le \frac{2}{n}$. Since $2/n \to 0$, the n-th term goes to 0 by the Squeeze Theorem.2. Geometric Series
Standard Theorem
A geometric series $\sum_{n=0}^{\infty} ar^n$ has a constant ratio $r$ between successive terms.
Converges if $|r| < 1$, diverges if $|r| \ge 1$.
If it converges, its exact sum is $S = \frac{a_1}{1 - r}$, where $a_1$ is the first term.
Common Mistake
Getting the first term wrong. Always plug in the starting index (e.g., $n=0$ or $n=1$) to find $a_1$ before using the sum formula.
Worked Examples
$\sum_{n=0}^{\infty} (1/2)^n$
$r = 1/2$. $|1/2| < 1$. First term $a_1 = (1/2)^0 = 1$.
Sum = $1 / (1 - 1/2) = 2$$\sum_{n=1}^{\infty} 3(2/3)^{n-1}$
$r = 2/3$. $|2/3| < 1$. First term $a_1 = 3(2/3)^0 = 3$.
Sum = $3 / (1 - 2/3) = 9$$\sum_{n=0}^{\infty} (-1/3)^n$
$r = -1/3$. $|-1/3| < 1$. First term $a_1 = 1$.
Sum = $1 / (1 - (-1/3)) = 3/4$$\sum_{n=2}^{\infty} 5(1/4)^n$
Missing first two terms! First term $a_1 = 5(1/4)^2 = 5/16$.
Sum = $(5/16) / (1 - 1/4) = 5/12$$\sum_{n=1}^{\infty} 7(-1/2)^n$
$r = -1/2$. First term $a_1 = 7(-1/2)^1 = -7/2$.
Sum = $(-7/2) / (1 - (-1/2)) = -7/3$$\sum_{n=0}^{\infty} 2^n$
$r = 2$. $|2| \ge 1$.
$\sum_{n=1}^{\infty} (-1)^n$
$r = -1$. $|-1| \ge 1$.
$\sum_{n=0}^{\infty} (5/4)^n$
$r = 5/4 = 1.25$. $|5/4| \ge 1$.
$\sum_{n=3}^{\infty} 6(1/5)^{n-3}$
Index shifted. First term is $6(1/5)^0 = 6$.
Sum = $6 / (1 - 1/5) = 30/4 = 15/2$$\sum_{n=0}^{\infty} -4(3/10)^n$
$r = 3/10$. First term $a_1 = -4$.
Sum = $-4 / (1 - 3/10) = -40/7$3. The p-Series Test
Standard Theorem
The series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if $p > 1$, and diverges if $p \le 1$.
Common Mistake
Mixing up $p>1$ with Ratio Test's $L<1$. For p-series, bigger powers in the denominator mean terms shrink faster, causing CONVERGENCE when $p>1$.
Worked Examples
$\sum \frac{1}{n^2}$
$p=2$.
Since 2 > 1, it converges.$\sum \frac{1}{n^{3/2}}$
$p=1.5$.
Since 1.5 > 1, it converges.$\sum \frac{1}{n^{1.01}}$
$p=1.01$.
Barely, but since 1.01 > 1, converges.$\sum \frac{1}{n}$
The Harmonic series. Equivalent to $p=1$.
Since 1 ≤ 1, it diverges.$\sum \frac{1}{\sqrt{n}}$
Rewritten as $\sum 1/n^{1/2}$. $p=0.5$.
Since 0.5 ≤ 1, it diverges.$\sum \frac{1}{n^{0.9}}$
$p=0.9$.
Since 0.9 ≤ 1, it diverges.$\sum \frac{5}{n^4}$
Constant multiple of a p-series where $p=4$.
Since 4 > 1, underlying converges.$\sum \frac{1}{n^{7/3}}$
$p=7/3 \approx 2.33$.
Since 7/3 > 1, it converges.$\sum \frac{1}{n^{2/3}}$
$p=2/3 \approx 0.67$.
Since 2/3 ≤ 1, it diverges.$\sum \frac{1}{(n\pi)^2}$
Expand to $\frac{1}{\pi^2} \sum \frac{1}{n^2}$. $p=2$.
Since 2 > 1, it converges.4. The Integral Test
Standard Theorem
Let $f(x)$ be continuous, positive, and decreasing for $x \ge 1$, and let $a_n = f(n)$.
$\sum_{n=1}^{\infty} a_n$ and $\int_{1}^{\infty} f(x) dx$ either BOTH converge or BOTH diverge.
Common Mistake
Forgetting to explicitly verify the 3 conditions: Continuous, Positive, Decreasing on the integration interval. A loose statement costs points on AP FRQs.
Worked Examples
$\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2}$
Let $f(x) = \frac{1}{x(\ln x)^2}$. For $x \ge 2$, $f$ is continuous, positive, and decreasing. Use $u$-substitution $u = \ln x, du = \frac{1}{x}dx$.
$\int_2^\infty u^{-2} du = \lim_{b\to\infty} \left[ -\frac{1}{\ln x} \right]_2^b = 0 - \left( -\frac{1}{\ln 2} \right) = \frac{1}{\ln 2}$The improper integral converges.
$\sum_{n=2}^{\infty} \frac{1}{n \ln n}$
Let $f(x) = \frac{1}{x \ln x}$. Substitution $u = \ln x$.
$\int_2^\infty \frac{du}{u} = \lim_{b\to\infty} [\ln |\ln x|]_2^b = \infty$The improper integral diverges.
$\sum_{n=3}^{\infty} \frac{1}{n \ln n \ln(\ln n)}$
Let $u = \ln(\ln x)$. Then $du = \frac{1}{x \ln x} dx$. Start at $n=3$ so that $\ln(\ln 3) > 0$.
$\int \frac{1}{u} du = [\ln|\ln(\ln x)|]_3^\infty = \infty$$\sum_{n=3}^{\infty} \frac{1}{n(\ln n)(\ln(\ln n))^2}$
Notice the start at $n=3$ to ensure positivity of logs. Let $u = \ln(\ln x)$.
$\int_3^\infty u^{-2} du = \lim_{b\to\infty} \left[-\frac{1}{\ln(\ln x)}\right]_3^b = 0 - \left(-\frac{1}{\ln(\ln 3)}\right) = \frac{1}{\ln(\ln 3)}$$\sum_{n=1}^{\infty} \frac{1}{n^2+1}$
Let $f(x) = \frac{1}{x^2+1}$. Classic inverse tangent integral.
$\int_1^\infty \frac{dx}{x^2+1} = \lim_{b\to\infty} [\arctan(x)]_1^b = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$$\sum_{n=1}^{\infty} \frac{1}{n^2+4}$
Use arctan integral rule: $\int \frac{dx}{x^2+a^2} = \frac{1}{a} \arctan(\frac{x}{a})$.
$\lim_{b\to\infty} \left[\frac{1}{2} \arctan \left(\frac{x}{2}\right)\right]_1^b = \frac{1}{2}\left(\frac{\pi}{2} - \arctan(1/2)\right)$$\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^3}$
Let $u = \ln x$.
$\int u^{-3} du = \lim_{b\to\infty} \left[ -\frac{1}{2(\ln x)^2} \right]_2^b = \frac{1}{2(\ln 2)^2}$$\sum_{n=2}^{\infty} \frac{\ln n}{n^2}$
Let $f(x) = \frac{\ln x}{x^2}$. Continuous, positive for $x \ge 2$. Decreasing eventually (derive $f'(x) = \frac{1-2\ln x}{x^3} < 0$ for $x > \sqrt{e}$). Use integration by parts.
$\int_2^\infty x^{-2} \ln(x) dx \to \text{Finite value}$$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^2+1}}$
Let $f(x) = \frac{1}{\sqrt{x^2+1}}$. Integral evaluates to $\ln|x+\sqrt{x^2+1}|$.
$\lim_{b\to\infty} \ln|b+\sqrt{b^2+1}| - \ln(1+\sqrt{2}) = \infty$$\sum_{n=1}^{\infty} \frac{1}{n+1}$
Let $f(x) = \frac{1}{x+1}$.
$\int_1^\infty \frac{dx}{x+1} = \lim_{b\to\infty} [\ln(x+1)]_1^b = \infty$5. Comparison Tests
Direct Comparison Test (DCT)
Standard Theorem
Let $0 \le a_n \le b_n$ for all $n$ (or at least for $n \ge M$).
If the "larger" series $\sum b_n$ converges, the "smaller" series $\sum a_n$ converges.
If the "smaller" series $\sum a_n$ diverges, the "larger" series $\sum b_n$ diverges.
Common Mistake
Getting the inequality backward. Knowing that $a_n < b_n$ and $\sum b_n$ diverges tells you absolutely nothing about $\sum a_n$. The inequalities must go in the correct direction!
Worked Examples (DCT)
$\sum \frac{1}{n^2+1}$
Compare to $\frac{1}{n^2}$. $n^2+1 > n^2 \implies \frac{1}{n^2+1} < \frac{1}{n^2}$.
$\sum 1/n^2$ converges (p=2>1). Smaller than convergent.$\sum \frac{1}{n^2+n}$
Compare to $\frac{1}{n^2}$. $n^2+n > n^2 \implies \frac{1}{n^2+n} < \frac{1}{n^2}$.
$\sum 1/n^2$ converges (p=2>1). Smaller than convergent.$\sum \frac{1}{n^3+5}$
Compare to $\frac{1}{n^3}$. $n^3+5 > n^3 \implies \frac{1}{n^3+5} < \frac{1}{n^3}$.
$\sum 1/n^3$ converges (p=3>1). Smaller than convergent.$\sum \frac{1}{\sqrt{n}+1}$
For $n \ge 1, \sqrt{n}+1 \le 2\sqrt{n}$, so $\frac{1}{\sqrt{n}+1} \ge \frac{1}{2\sqrt{n}}$.
$\sum \frac{1}{2\sqrt{n}}$ diverges. Larger than divergent.$\sum_{n=3}^{\infty} \frac{1}{n-2}$
Compare to $\frac{1}{n}$. $n-2 < n \implies \frac{1}{n-2} > \frac{1}{n}$.
$\sum 1/n$ diverges. Larger than divergent.$\sum_{n=2}^{\infty} \frac{1}{n^2-n}$
Compare to $\frac{2}{n^2}$. $n^2-n \ge n^2/2 \implies \frac{1}{n^2-n} \le \frac{2}{n^2}$.
$\sum \frac{2}{n^2}$ converges. Smaller than convergent.$\sum \frac{n}{n^3+1}$
Compare to $\frac{n}{n^3} = \frac{1}{n^2}$. $\frac{n}{n^3+1} < \frac{1}{n^2}$.
Smaller than a convergent p-series.$\sum \frac{n+1}{n^2}$
Compare to $\frac{n}{n^2} = \frac{1}{n}$. $\frac{n+1}{n^2} > \frac{1}{n}$.
Larger than divergent harmonic series.$\sum \frac{1}{n^4+n}$
Compare to $\frac{1}{n^4}$. $n^4+n > n^4 \implies \frac{1}{n^4+n} < \frac{1}{n^4}$.
Smaller than a convergent p-series.$\sum \frac{1}{n^{1/3}+5}$
For large $n$, $\frac{1}{n^{1/3}+5} \ge \frac{1}{2n^{1/3}}$.
Larger than divergent p-series ($p=1/3$).Limit Comparison Test (LCT)
Standard Theorem
Let $a_n > 0$ and $b_n > 0$. If $\lim_{n \to \infty} \frac{a_n}{b_n} = L$ where $0 < L < \infty$:
Both series converge, or both series diverge. (They share the same behavior.)
Worked Examples (LCT)
$\sum \frac{2n+1}{n^2}$
Behaves like $2n/n^2 = 2/n$. Let $b_n = 1/n$.
$\lim \frac{(2n+1)/n^2}{1/n} = 2$. $0 < 2 < \infty$. $\sum 1/n$ diverges.$\sum \frac{1}{n^2-n}$
Behaves like $1/n^2$. Let $b_n = 1/n^2$.
$\lim \frac{1/(n^2-n)}{1/n^2} = 1$. $0 < 1 < \infty$. $\sum 1/n^2$ converges.$\sum \frac{3n^2+1}{n^3+2}$
Behaves like $3n^2/n^3 = 3/n$. Let $b_n = 1/n$.
$\lim \frac{(3n^2+1)/(n^3+2)}{1/n} = 3$. $0 < 3 < \infty$. $\sum 1/n$ div.$\sum \frac{n+4}{n^2+7}$
Behaves like $n/n^2 = 1/n$. Let $b_n = 1/n$.
$\lim \frac{(n+4)/(n^2+7)}{1/n} = 1$. $\sum 1/n$ diverges.$\sum \frac{5n^3-1}{2n^5+3}$
Behaves like $5n^3/2n^5 = (5/2)(1/n^2)$. Let $b_n = 1/n^2$.
$\lim \frac{(5n^3-1)/(2n^5+3)}{1/n^2} = 5/2$. $\sum 1/n^2$ converges.$\sum \frac{\sqrt{n^2+1}}{n^3+1}$
Behaves like $\sqrt{n^2}/n^3 = n/n^3 = 1/n^2$. Let $b_n = 1/n^2$.
$\lim \frac{\sqrt{n^2+1}/(n^3+1)}{1/n^2} = 1$. $\sum 1/n^2$ converges.$\sum \frac{n^2+5}{n^4+1}$
Behaves like $n^2/n^4 = 1/n^2$. Let $b_n = 1/n^2$.
$\lim \frac{(n^2+5)/(n^4+1)}{1/n^2} = 1$. $\sum 1/n^2$ converges.$\sum \frac{4n-1}{n^2+9}$
Behaves like $4n/n^2 = 4/n$. Let $b_n = 1/n$.
$\lim \frac{(4n-1)/(n^2+9)}{1/n} = 4$. $\sum 1/n$ diverges.$\sum \frac{\ln n}{n}$ compared with $\frac{1}{n}$
CAUTION: DO NOT USE LCT HERE.
$\lim \frac{(\ln n)/n}{1/n} = \lim \ln n = \infty$Because $L = \infty$, standard LCT fails.
$\sum \frac{n^2+\sin n}{3n^2+1}$
Behaves like $n^2/3n^2 = 1/3$.
LCT with $b_n=1$ yields $1/3$. $\sum 1$ div. Or $a_n \to 1/3 \neq 0$.6. The Alternating Series Test (AST)
Standard Theorem
Let $\sum (-1)^{n} b_n$ be an alternating series where $b_n \ge 0$. The series converges if:
1) $b_{n+1} \le b_n$ eventually (terms are decreasing in magnitude)
2) $\lim_{n \to \infty} b_n = 0$
Warning about signs
Alternating signs only guarantee convergence if the terms actually shrink to zero! If $b_n \to 1$, it STILL diverges by n-th term test.
Absolute vs. Conditional Preview:
If $\sum |a_n|$ converges, it converges absolutely. If $\sum a_n$ converges but $\sum |a_n|$ diverges, it converges conditionally.
Worked Examples
$\sum \frac{(-1)^{n+1}}{n}$
Alternating harmonic. $b_n = 1/n \to 0$ and $1/(n+1) < 1/n$.
But $\sum 1/n$ diverges. Thus, converges, but not absolutely.$\sum \frac{(-1)^n}{n^2}$
Absolute value series $\sum 1/n^2$ converges (p=2).
AST is valid but not needed; absolute convergence is stronger.$\sum \frac{(-1)^{n+1}}{\sqrt{n}}$
$b_n = 1/\sqrt{n} \to 0$ and is decreasing. AST applies.
However, absolute series $\sum 1/\sqrt{n}$ diverges (p=1/2).$\sum (-1)^n \frac{n}{n+1}$
$b_n = n/(n+1) \to 1 \neq 0$.
Terms fail to go to zero. Fails n-th term test.$\sum (-1)^{n+1} \frac{\ln n}{n}$
$b_n = (\ln n)/n \to 0$ via L'Hôpital. Decreasing for $n > e$.
Absolute series diverges via Integral Test. AST gives convergence.$\sum \frac{(-1)^n}{n+3}$
$b_n = 1/(n+3) \to 0$, decreasing. AST proves convergence.
Series without $(-1)^n$ acts like harmonic $1/n$ (divergence).$\sum (-1)^n \frac{n}{n^2+1}$
$b_n = n/(n^2+1) \to 0$, decreasing. AST proves convergence.
Absolute series divergent (LCT with $1/n$).$\sum \frac{(-1)^n}{2^n}$
Alternating geometric with $r = -1/2$.
Absolute series is geometric with $r=1/2$, which converges.$\sum (-1)^n (1+\frac{1}{n})$
AST requires $b_n \to 0$. Here $b_n = 1 + 1/n \to 1$.
The terms alternate between numbers close to $1$ and $-1$.$\sum (-1)^n \frac{1 + \sin(1/n)}{n}$
$b_n = \frac{1+\sin(1/n)}{n} \sim \frac{1+0}{n} = \frac{1}{n} \to 0$. AST converges.
Absolute series acts like $1/n$ (diverges).8. The Ratio & Root Tests
The Ratio Test
Standard Theorem
Let $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$.
If $L < 1$, the series converges absolutely.
If $L > 1$ (or $\infty$), the series diverges.
If $L = 1$, the test is inconclusive.
Worked Examples (Ratio Test)
$\sum \frac{n^3}{3^n}$
Limit of $|a_{n+1}/a_n|$: $\lim \left| \frac{(n+1)^3/3^{n+1}}{n^3/3^n} \right| = \lim \frac{(n+1)^3}{n^3} \cdot \frac{3^n}{3^{n+1}} = 1 \cdot \frac{1}{3} = \frac{1}{3}$
$1/3 < 1$. Converges absolutely.$\sum \frac{n!}{100^n}$
$\lim \left| \frac{(n+1)!/100^{n+1}}{n!/100^n} \right| = \lim \frac{n+1}{100} = \infty$
$\infty > 1$. Diverges. Factorials beat exponentials.$\sum \frac{2^n}{n!}$
$\lim \left| \frac{2^{n+1}/(n+1)!}{2^n/n!} \right| = \lim \frac{2}{n+1} = 0$
$0 < 1$. Converges absolutely.$\sum \frac{n^2}{(2n)!}$
$\lim \left| \frac{(n+1)^2}{(2n+2)!} \frac{(2n)!}{n^2} \right| = \lim \frac{(n+1)^2}{n^2 (2n+1)(2n+2)} = 0$
$0 < 1$. Converges absolutely.$\sum \frac{(-1)^n 3^n}{n^2 4^n}$
The limit of the ratio is $\lim \left| \frac{3^{n+1}/(n+1)^2 4^{n+1}}{3^n/n^2 4^n} \right| = \frac{3}{4} \lim \frac{n^2}{(n+1)^2} = \frac{3}{4}$
$3/4 < 1$. Converges absolutely.$\sum \frac{5^n}{n^5}$
$\lim \left| \frac{5^{n+1}/(n+1)^5}{5^n/n^5} \right| = 5 \lim \left(\frac{n}{n+1}\right)^5 = 5(1) = 5$
$5 > 1$. Diverges.$\sum \frac{n!}{(2n)!}$
$\lim \frac{(n+1)!}{(2n+2)!} \cdot \frac{(2n)!}{n!} = \lim \frac{n+1}{(2n+2)(2n+1)} = 0$
$0 < 1$. Converges absolutely.$\sum \frac{e^n}{n!}$
$\lim \frac{e^{n+1}/(n+1)!}{e^n/n!} = \lim \frac{e}{n+1} = 0$
$0 < 1$. Converges absolutely.$\sum \frac{n^n}{n!}$
$\lim \frac{(n+1)^{n+1}/(n+1)!}{n^n/n!} = \lim \left(\frac{n+1}{n}\right)^n = e$
$e \approx 2.718 > 1$. Diverges.$\sum \frac{(2n)!}{n!(n!)}$
$\lim \frac{(2n+2)!}{((n+1)!)^2} \frac{(n!)^2}{(2n)!} = \lim \frac{(2n+2)(2n+1)}{(n+1)^2} = 4$
$4 > 1$. Diverges.The Root Test
Standard Theorem
Let $L = \lim_{n \to \infty} \sqrt[n]{|a_n|}$.
If $L < 1$, the series converges absolutely.
If $L > 1$ (or $\infty$), the series diverges.
If $L = 1$, the test is inconclusive.
Worked Examples (Root Test)
$\sum \left( \frac{2n}{3n+1} \right)^n$
Evaluate limit of the nth root: $\lim \sqrt[n]{|a_n|} = \lim \frac{2n}{3n+1} = \frac{2}{3}$.
$2/3 < 1$. Converges absolutely.$\sum \left( 1 - \frac{1}{n} \right)^{n^2}$
$\lim \left( (1 - 1/n)^{n^2} \right)^{1/n} = \lim (1 - 1/n)^n = e^{-1} = 1/e$
$1/e \approx 0.368 < 1$. Converges absolutely.$\sum \left( \frac{5n+2}{n+7} \right)^{2n}$
Take the n-th root of the absolute terms: $\sqrt[n]{|a_n|} = \left( \frac{5n+2}{n+7} \right)^2$.
Then take the limit: $\lim \left( \frac{5n+2}{n+7} \right)^2 = 5^2 = 25$.
$\sum \left( \frac{2n+5}{2n+1} \right)^n$
Root Test gives $\lim \frac{2n+5}{2n+1} = 1$, which is INCONCLUSIVE.
Because $L=1$, the Root Test is inconclusive. We must evaluate the terms themselves for the n-th Term Test:
$\lim_{n \to \infty} \left( \frac{2n+5}{2n+1} \right)^n = \lim_{n \to \infty} \left( 1 + \frac{4}{2n+1} \right)^n$Since $n \cdot \frac{4}{2n+1} \to 2$, this standard limit approaches $e^2$.
Since the terms tend to $e^2 \neq 0$, the series diverges by the n-th Term Test.
$\sum \frac{3^n}{n^n}$
$\lim \sqrt[n]{3^n/n^n} = \lim \frac{3}{n} = 0$
$0 < 1$. Converges absolutely.7. Absolute vs. Conditional Convergence
When dealing with series that have both positive and negative terms, we classify their convergence into two strict types:
Absolute Convergence
A series $\sum a_n$ converges absolutely if the series of its absolute values $\sum |a_n|$ converges.
Example: $\sum \frac{(-1)^n}{n^2}$ converges absolutely because $\sum \frac{1}{n^2}$ converges (p=2).
Conditional Convergence
A series $\sum a_n$ converges conditionally if the original series $\sum a_n$ converges, BUT the absolute value series $\sum |a_n|$ diverges.
9. Power Series, Radius, and Interval
Definition & Procedure
A Power Series is an infinite polynomial centered at $x=c$:
How to find the Interval of Convergence:
- Apply the Ratio Test: $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1$.
- Solve the resulting inequality for $|x - c| < R$. $R$ is the Radius.
- Write the open interval $(c-R, c+R)$.
- MANDATORY: Test individual endpoints $x = c-R$ and $x = c+R$ by plugging them into the original series and running standard tests.
Worked Examples (Intervals of Convergence)
$\sum_{n=1}^{\infty} x^n$
$\lim |x^{n+1}/x^n| = |x| < 1$Endpoints: $x=1 \implies \sum 1^n$ (Div), $x=-1 \implies \sum (-1)^n$ (Div).
$\sum_{n=1}^{\infty} \frac{(x-2)^n}{n}$
$\lim \left| \frac{(x-2)^{n+1}/(n+1)}{(x-2)^n/n} \right| = |x-2| < 1$Endpoints $x=1, x=3$.
$x=3 \implies \sum 1/n$ (Div p-series).
$x=1 \implies \sum (-1)^n/n$ (Conv AST).
$\sum_{n=1}^{\infty} \frac{n(x+1)^n}{3^n}$
$\lim \frac{|x+1|}{3} < 1 \implies |x+1| < 3$Endpoints $x=-4, 2$. Both yield $n(\pm 1)^n$, failing n-th term test.
$\sum_{n=0}^{\infty} \frac{(x-4)^n}{n!}$
$\lim \frac{|x-4|}{n+1} = 0 < 1$Limit is 0 for ANY $x$. Converges everywhere.
$\sum_{n=1}^{\infty} \frac{n! x^n}{5^n}$
$\lim \left| \frac{(n+1)! x^{n+1}/5^{n+1}}{n! x^n/5^n} \right| = \lim \frac{(n+1)|x|}{5} = \infty$If $x \neq 0$, the limit is $\infty$. Thus it only converges at the center.
$\sum_{n=1}^{\infty} \frac{(x-1)^n}{n^2}$
$\lim |(x-1)| < 1$. Endpoints $0, 2$.$x=2 \implies \sum 1/n^2$ (Conv).
$x=0 \implies \sum (-1)^n/n^2$ (Conv absolutely).
$\sum_{n=1}^{\infty} \frac{(x+3)^n}{n 2^n}$
$\lim \frac{|x+3|}{2} < 1 \implies |x+3| < 2$. Endpoints $-5, -1$.$x=-1 \implies \sum 1/n$ (Div).
$x=-5 \implies \sum (-1)^n/n$ (Conv AST).
$\sum_{n=1}^{\infty} n(x-5)^n$
$\lim |x-5| < 1$. Endpoints 4, 6.$x=6 \implies \sum n$ (Div).
$x=4 \implies \sum n(-1)^n$ (Div).
10. Taylor & Maclaurin Series
A Taylor Series is a power series that exactly matches a function $f(x)$ at $x=c$ and all its derivatives:
A Maclaurin Series is just a Taylor series centered at $c=0$.
Common Standard Series (Must Memorize)
| Function | Maclaurin Series | Interval |
|---|---|---|
| $e^x$ | $\sum \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$ | $(-\infty, \infty)$ |
| $\sin x$ | $\sum \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$ | $(-\infty, \infty)$ |
| $\cos x$ | $\sum \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$ | $(-\infty, \infty)$ |
| $\frac{1}{1-x}$ | $\sum x^n = 1 + x + x^2 + x^3 + \dots$ | $(-1, 1)$ |
| $\ln(1+x)$ | $\sum \frac{(-1)^{n-1} x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$ | $(-1, 1]$ |
Building New Series
Instead of computing derivatives mechanically, substitute into known Maclaurin series:
Find Maclaurin for $f(x) = x^2 e^{3x}$
Sub $3x$ into $e^x$: $e^{3x} = \sum \frac{(3x)^n}{n!}$Multiply by $x^2$: $\sum \frac{3^n x^{n+2}}{n!}$
Find Maclaurin for $f(x) = \int \sin(x^2) dx$
Sub $x^2$ into $\sin x$: $\sum \frac{(-1)^n (x^2)^{2n+1}}{(2n+1)!} = \sum \frac{(-1)^n x^{4n+2}}{(2n+1)!}$Integrate term-by-term: $\sum \frac{(-1)^n x^{4n+3}}{(4n+3)(2n+1)!} + C$
If no initial condition is given, $C$ remains an arbitrary constant. Only solve for $C$ if an initial value is specified.
11. Error Bounds
Alternating Series Error Bound
If an alternating series satisfies the AST, the error by stopping at $n$ terms is bounded by the magnitude of the very next term.
Example: Stopping conditions
How many terms of $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$ are needed to be within 0.01 of the true sum?
Need $|a_{n+1}| \le 0.01 \implies \frac{1}{(n+1)^2} \le 0.01 \implies (n+1)^2 \ge 100 \implies n+1 \ge 10 \implies n \ge 9$.Lagrange Error Bound
For ANY Taylor polynomial approximation, the error remainder $R_n(x)$ is bounded by:
Where $M$ is the MAX value of $|f^{(n+1)}(z)|$ between the center $c$ and $x$.
Example: Bounding $P_2(x)$ for $\cos x$ at $x=0.1$
$n=2$, $c=0$, $x=0.1$. Next derivative $f'''(z) = \sin(z)$. Max of $|\sin z|$ is $\le 1$. Thus $M=1$.
$|R_2| \le \frac{1}{3!} |0.1 - 0|^3 = \frac{0.001}{6}$.12. Test Selection Strategy (The Flowchart)
Stuck on a series? Try this strong decision path in this order first:
-
1
n-th Term Test
Just look at the terms as $n \to \infty$. Do they fail to go to zero? If so, DIVERGES. YOU ARE DONE.
-
2
Is it a special series?
Look for $a r^n$ (Geometric, check $|r|<1$) or $1/n^p$ (p-Series, check $p>1$).
-
3
Does it have $(-1)^n$?
Use Alternating Series Test (AST). Verify terms shrink to 0.
Follow up: Immediately check for Absolute vs Conditional convergence by analyzing the series without the alternating sign. -
4
Does it resemble a p-series, but is messy?
Use a Comparison Test.
- Limit Comparison Test (LCT): Best for rational functions (polynomial over polynomial). Take the highest degree $n$ terms.
- Direct Comparison Test (DCT): Best when LCT fails (e.g., when terms have $\sin^2 n$, or bounded oscillations, or annoying logarithms where LCT limits go to 0 or $\infty$).
-
5
Does it have factorials or $c^n$?
Use Ratio Test for factorials and exponentials combinations.
Use Root Test if the ENTIRE expression is wrapped in a power of $n$ (like $(\dots)^n$ or $(\dots)^{n^2}$).
For Power Series:Always use the Ratio Test (or Root Test) first to find the open radius of convergence.
Then, test the individual border endpoints separately by plugging them into the original series.
-
6
Is it something easily integrable?
Use Integral Test. E.g., $1/(x \ln x)$ or $xe^{-x^2}$. Check positive, continuous, decreasing.
AP Exam Survival Traps to Avoid
- Ratio/Root $L=1$: This does not mean it diverges. It means you must try another test.
- Power Series Endpoints: ALWAYS test endpoints when finding the Interval of Convergence. They are guaranteed points on the FRQ!
- Integrals as Series: Remember the $+ C$ when integrating a Taylor series to form a new series, and solve for $C$ by plugging in the series center point.