Chapter 10: Infinite Sequences and Series BC ONLY
Infinite series is the single most important topic that separates AP Calculus BC from AB. This chapter covers everything you need to know: sequences, convergence tests, power series, Taylor and Maclaurin series, and Lagrange error bounds. Roughly 17–20% of the BC exam focuses on series, making this chapter essential for a high score.
Table of Contents
10.1 Sequences
Think of a sequence as an infinite list: $a_1, a_2, a_3, \ldots$ Each term is determined by plugging $n$ into a formula. For example, if $a_n = \frac{1}{n}$, the sequence is $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots$
Convergence and Divergence
In practical terms, a sequence converges if its terms approach a fixed number as $n$ grows. To evaluate $\lim_{n \to \infty} a_n$, you can use the same techniques as limits of functions: direct substitution, L'Hopital's Rule (by treating $n$ as a continuous variable $x$), and the Squeeze Theorem.
Bounded and Monotonic Sequences
A sequence $\{a_n\}$ is:
- Bounded above if there exists $M$ such that $a_n \leq M$ for all $n$
- Bounded below if there exists $m$ such that $a_n \geq m$ for all $n$
- Monotonically increasing if $a_{n+1} \geq a_n$ for all $n$
- Monotonically decreasing if $a_{n+1} \leq a_n$ for all $n$
This theorem is powerful because it guarantees convergence without requiring you to find the limit explicitly. If you can show a sequence is increasing and bounded above (or decreasing and bounded below), it must converge.
Find $\lim_{n \to \infty} \frac{3n^2 + 1}{5n^2 - 2n}$.
Solution: Divide numerator and denominator by $n^2$:
$$\lim_{n \to \infty} \frac{3 + \frac{1}{n^2}}{5 - \frac{2}{n}} = \frac{3 + 0}{5 - 0} = \frac{3}{5}$$The sequence converges to $\frac{3}{5}$.
Find $\lim_{n \to \infty} \frac{n!}{n^n}$.
Solution: Write out the ratio:
$$\frac{n!}{n^n} = \frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{n}{n}$$Every factor $\frac{k}{n} \leq 1$, and the first factor is $\frac{1}{n} \to 0$. Since $0 \leq \frac{n!}{n^n} \leq \frac{1}{n}$ and $\frac{1}{n} \to 0$, by the Squeeze Theorem:
$$\lim_{n \to \infty} \frac{n!}{n^n} = 0$$The sequence converges to $0$.
10.2 Series and Convergence
Geometric Series
The geometric series is one of the most important series on the AP exam. You must be able to identify a geometric series, determine if it converges, and compute its sum. Watch for series that can be rewritten in geometric form through algebraic manipulation.
Find the sum of $\displaystyle\sum_{n=1}^{\infty} \frac{3}{4^n}$.
Solution: Rewrite as $\displaystyle\sum_{n=1}^{\infty} 3 \cdot \left(\frac{1}{4}\right)^n = 3 \cdot \sum_{n=1}^{\infty} \left(\frac{1}{4}\right)^n$.
This is a geometric series starting at $n=1$ with $a = \frac{1}{4}$ and $r = \frac{1}{4}$. Since the sum starting at $n=0$ is $\frac{1}{1 - 1/4} = \frac{4}{3}$, the sum starting at $n=1$ is $\frac{4}{3} - 1 = \frac{1}{3}$.
Alternatively, use the formula with first term $\frac{3}{4}$ and ratio $\frac{1}{4}$:
$$\frac{3/4}{1 - 1/4} = \frac{3/4}{3/4} = 1$$The sum is $1$.
Geometric series partial sums: S_n = Σ r^k from k=0 to n. When |r| < 1, the partial sums converge to 1/(1-r). Adjust r to see convergence vs divergence.
Telescoping Series
A telescoping series is one where most terms cancel in the partial sum. The classic form uses partial fractions:
Find the sum of $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$.
Solution: Use partial fractions: $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.
Write out the partial sum:
$$S_n = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right)$$After cancellation: $S_n = 1 - \frac{1}{n+1}$. Therefore:
$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right) = 1$$Divergence Test (nth Term Test)
Important caveat: If $\lim_{n \to \infty} a_n = 0$, the series may or may not converge. The harmonic series $\sum \frac{1}{n}$ diverges even though $\frac{1}{n} \to 0$. The Divergence Test can only prove divergence, never convergence.
Does $\displaystyle\sum_{n=1}^{\infty} \frac{n}{2n+3}$ converge or diverge?
Solution: Check the limit: $\lim_{n \to \infty} \frac{n}{2n+3} = \frac{1}{2} \neq 0$.
Since the terms do not approach zero, the series diverges by the Divergence Test.
10.3 Convergence Tests
This section covers the major convergence tests you need for the AP exam. Choosing the right test is a critical skill. After presenting each test, we provide a comprehensive summary table.
Integral Test
The Integral Test does not give you the sum of the series—it only tells you whether the series converges or diverges. The value of the integral is generally not equal to the sum of the series.
p-Series Test
Key cases to memorize: $\sum \frac{1}{n}$ diverges ($p = 1$, the harmonic series), $\sum \frac{1}{n^2}$ converges ($p = 2$), and $\sum \frac{1}{\sqrt{n}}$ diverges ($p = \frac{1}{2}$).
Comparison Tests
- If $\sum b_n$ converges, then $\sum a_n$ converges.
- If $\sum a_n$ diverges, then $\sum b_n$ diverges.
The Limit Comparison Test is often easier to use than Direct Comparison because you don't need to establish an inequality—you just compute a limit. Compare your series to a known p-series or geometric series.
Alternating Series Test
- $b_n > 0$ for all $n$
- $b_{n+1} \leq b_n$ for all $n$ (decreasing)
- $\lim_{n \to \infty} b_n = 0$
Ratio Test
- If $L < 1$, the series converges absolutely.
- If $L > 1$ (or $L = \infty$), the series diverges.
- If $L = 1$, the test is inconclusive.
The Ratio Test is the go-to test for series involving factorials, exponentials, or products of both. It is also the primary tool for finding the radius of convergence of a power series.
Root Test
- If $L < 1$, the series converges absolutely.
- If $L > 1$, the series diverges.
- If $L = 1$, the test is inconclusive.
The Root Test is useful when $a_n$ has an $n$th power form like $(c_n)^n$. It appears less frequently on the AP exam than the Ratio Test, but it is good to know.
Convergence Test Summary Table
| Test | When to Use | Conclusion |
|---|---|---|
| Divergence Test | Always try first: check if $a_n \to 0$ | If $\lim a_n \neq 0$: diverges. If $= 0$: inconclusive |
| Geometric Series | Series has the form $\sum ar^n$ | $|r|<1$: converges to $\frac{a}{1-r}$. $|r|\geq 1$: diverges |
| p-Series | Series has the form $\sum \frac{1}{n^p}$ | $p > 1$: converges. $p \leq 1$: diverges |
| Integral Test | $f(n) = a_n$ is positive, continuous, decreasing | $\int_1^\infty f(x)\,dx$ converges $\Leftrightarrow$ $\sum a_n$ converges |
| Direct Comparison | Can bound $a_n$ above/below by a known series | $a_n \leq b_n$ and $\sum b_n$ conv. $\Rightarrow$ $\sum a_n$ conv. |
| Limit Comparison | $a_n$ resembles a known series for large $n$ | If $\lim \frac{a_n}{b_n} = L \in (0,\infty)$: both behave the same |
| Alternating Series | Series has $(-1)^n$ factor with decreasing terms | $b_n$ decreasing, $b_n \to 0$: converges |
| Ratio Test | Factorials, exponentials, or $n!$ and $a^n$ terms | $L < 1$: conv. abs. $L > 1$: div. $L = 1$: inconclusive |
| Root Test | Terms with $n$th power form $(c_n)^n$ | Same as Ratio Test conclusions |
Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}$ converges or diverges.
Solution: Let $f(x) = \frac{1}{x^2+1}$. This function is positive, continuous, and decreasing on $[1, \infty)$.
$$\int_1^{\infty} \frac{1}{x^2+1}\,dx = \lim_{b \to \infty} \left[\arctan x\right]_1^b = \lim_{b \to \infty} \left(\arctan b - \arctan 1\right) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$$Since the integral converges, the series $\sum \frac{1}{n^2+1}$ converges by the Integral Test.
Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{n}{n^3 + 4}$ converges or diverges.
Solution: For large $n$, $\frac{n}{n^3+4} \approx \frac{n}{n^3} = \frac{1}{n^2}$. Compare with $b_n = \frac{1}{n^2}$:
$$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{n/(n^3+4)}{1/n^2} = \lim_{n \to \infty} \frac{n^3}{n^3+4} = 1$$Since $L = 1 \in (0, \infty)$ and $\sum \frac{1}{n^2}$ converges (p-series, $p=2$), the series $\sum \frac{n}{n^3+4}$ converges by the Limit Comparison Test.
Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ converges or diverges.
Solution: This is the alternating harmonic series. Let $b_n = \frac{1}{n}$.
- $b_n > 0$ for all $n$. ✓
- $b_{n+1} = \frac{1}{n+1} < \frac{1}{n} = b_n$ (decreasing). ✓
- $\lim_{n \to \infty} b_n = 0$. ✓
All three conditions are met, so the alternating harmonic series converges by the Alternating Series Test. (Its sum is $\ln 2$.)
Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{n!}{3^n}$ converges or diverges.
Solution: Apply the Ratio Test:
$$L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{(n+1)!}{3^{n+1}} \cdot \frac{3^n}{n!} = \lim_{n \to \infty} \frac{n+1}{3} = \infty$$Since $L = \infty > 1$, the series diverges by the Ratio Test.
10.4 Absolute and Conditional Convergence
- A series $\sum a_n$ converges absolutely if $\sum |a_n|$ converges.
- A series $\sum a_n$ converges conditionally if $\sum a_n$ converges but $\sum |a_n|$ diverges.
The converse is not true. The alternating harmonic series $\sum \frac{(-1)^{n+1}}{n}$ converges, but $\sum \frac{1}{n}$ (the harmonic series) diverges. So the alternating harmonic series converges conditionally, not absolutely.
The distinction matters because absolutely convergent series have stronger properties: you can rearrange terms without changing the sum. Conditionally convergent series can be rearranged to converge to any number (the Riemann Rearrangement Theorem).
Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$ converges absolutely, converges conditionally, or diverges.
Solution: Check absolute convergence first. Consider $\sum |a_n| = \sum \frac{1}{n^2}$.
This is a p-series with $p = 2 > 1$, so $\sum \frac{1}{n^2}$ converges.
Since $\sum |a_n|$ converges, the original series converges absolutely.
10.5 Power Series
Radius and Interval of Convergence
Every power series has a radius of convergence $R$ such that:
- The series converges absolutely for $|x - a| < R$
- The series diverges for $|x - a| > R$
- At $|x - a| = R$ (the endpoints), you must check each endpoint separately
The radius $R$ can be $0$ (converges only at $x=a$), a positive number, or $\infty$ (converges for all $x$). The interval of convergence includes all $x$ for which the series converges, and it may or may not include the endpoints.
Finding the Radius of Convergence
Use the Ratio Test. For $\sum c_n(x-a)^n$, compute:
$$L = \lim_{n \to \infty} \left|\frac{c_{n+1}}{c_n}\right| \cdot |x - a|$$The series converges when $L < 1$, which gives $|x-a| < \frac{1}{\lim |c_{n+1}/c_n|}$. Therefore $R = \frac{1}{\lim |c_{n+1}/c_n|}$.
Find the radius and interval of convergence for $\displaystyle\sum_{n=1}^{\infty} \frac{x^n}{n}$.
Solution: Apply the Ratio Test:
$$L = \lim_{n \to \infty} \left|\frac{x^{n+1}/(n+1)}{x^n/n}\right| = \lim_{n \to \infty} |x| \cdot \frac{n}{n+1} = |x|$$Converges when $|x| < 1$, so $R = 1$. Now check endpoints:
- $x = 1$: $\sum \frac{1}{n}$ is the harmonic series, which diverges.
- $x = -1$: $\sum \frac{(-1)^n}{n}$ is the alternating harmonic series, which converges (by AST).
Interval of convergence: $[-1, 1)$.
Find the radius of convergence for $\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!}$.
Solution: Apply the Ratio Test:
$$L = \lim_{n \to \infty} \left|\frac{x^{n+1}/(n+1)!}{x^n/n!}\right| = \lim_{n \to \infty} \frac{|x|}{n+1} = 0$$Since $L = 0 < 1$ for all $x$, the series converges for all real numbers. The radius of convergence is $R = \infty$, and the interval is $(-\infty, \infty)$.
(This is the Maclaurin series for $e^x$.)
Find the radius and interval of convergence for $\displaystyle\sum_{n=0}^{\infty} \frac{(x-3)^n}{2^n}$.
Solution: This is a geometric series with ratio $r = \frac{x-3}{2}$. It converges when $\left|\frac{x-3}{2}\right| < 1$, i.e., $|x-3| < 2$.
Radius of convergence: $R = 2$. The interval before endpoint checks: $(1, 5)$.
- $x = 1$: $\sum \frac{(-2)^n}{2^n} = \sum (-1)^n$ diverges.
- $x = 5$: $\sum \frac{2^n}{2^n} = \sum 1$ diverges.
Interval of convergence: $(1, 5)$.
10.6 Taylor and Maclaurin Series
Must-Know Maclaurin Series
| Function | Maclaurin Series | Interval |
|---|---|---|
| $e^x$ | $\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$ | $(-\infty, \infty)$ |
| $\sin x$ | $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$ | $(-\infty, \infty)$ |
| $\cos x$ | $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$ | $(-\infty, \infty)$ |
| $\dfrac{1}{1-x}$ | $\displaystyle\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots$ | $(-1, 1)$ |
| $\ln(1+x)$ | $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$ | $(-1, 1]$ |
| $\arctan x$ | $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots$ | $[-1, 1]$ |
Taylor polynomial approximation of sin(x) centered at 0. Increase the degree n to see how the polynomial matches the function more closely.
Operations on Power Series
You can create new Taylor series from known ones using these operations:
- Substitution: Replace $x$ with an expression. Example: The series for $e^{-x^2}$ is obtained by substituting $-x^2$ for $x$ in the series for $e^x$.
- Differentiation: Differentiate term by term within the interval of convergence. Example: Since $\frac{1}{1-x} = \sum x^n$, differentiating gives $\frac{1}{(1-x)^2} = \sum nx^{n-1}$.
- Integration: Integrate term by term. Example: Since $\frac{1}{1+x} = \sum (-1)^n x^n$, integrating gives $\ln(1+x) = \sum \frac{(-1)^n x^{n+1}}{n+1}$.
- Multiplication: Multiply two known series (Cauchy product), or multiply a series by a polynomial.
Find the Maclaurin series for $e^{-x^2}$.
Solution: Start with $e^u = \sum_{n=0}^{\infty} \frac{u^n}{n!}$ and substitute $u = -x^2$:
$$e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{n!} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots$$This series converges for all $x$ (since the series for $e^u$ converges for all $u$).
Find the Taylor series for $f(x) = \ln x$ centered at $a = 1$, up to the general term.
Solution: Compute derivatives at $a = 1$:
- $f(x) = \ln x \Rightarrow f(1) = 0$
- $f'(x) = \frac{1}{x} \Rightarrow f'(1) = 1$
- $f''(x) = -\frac{1}{x^2} \Rightarrow f''(1) = -1$
- $f'''(x) = \frac{2}{x^3} \Rightarrow f'''(1) = 2$
- $f^{(n)}(x) = \frac{(-1)^{n-1}(n-1)!}{x^n} \Rightarrow f^{(n)}(1) = (-1)^{n-1}(n-1)!$
The Taylor series is:
$$\ln x = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(n-1)!}{n!}(x-1)^n = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}(x-1)^n$$Which equals $(x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \cdots$, valid for $0 < x \leq 2$.
Use a known Maclaurin series to find $\displaystyle\int_0^1 \frac{\sin x}{x}\,dx$ as an infinite series.
Solution: Since $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$, we have:
$$\frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n+1)!}$$Integrating term by term from $0$ to $1$:
$$\int_0^1 \frac{\sin x}{x}\,dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)(2n+1)!} = 1 - \frac{1}{3 \cdot 3!} + \frac{1}{5 \cdot 5!} - \frac{1}{7 \cdot 7!} + \cdots$$This is an alternating series, so partial sums give accurate approximations with known error bounds.
10.7 Taylor Polynomial Error Bound (Lagrange)
When we approximate a function $f(x)$ using a Taylor polynomial $P_n(x)$, the error (or remainder) is $R_n(x) = f(x) - P_n(x)$. The Lagrange error bound gives us a way to control this error.
The key challenge is finding $M$, the maximum value of $|f^{(n+1)}|$ on the relevant interval. For many common functions, this is straightforward:
- For $\sin x$ and $\cos x$: all derivatives are bounded by $1$, so $M = 1$.
- For $e^x$ on $[0, b]$: the $(n+1)$th derivative is $e^x$, so $M = e^b$.
The Alternating Series Error Bound is often simpler to apply than Lagrange. If the series is alternating and meets the AST conditions, the error is no more than the absolute value of the next term.
Use the third-degree Taylor polynomial for $\sin x$ centered at $0$ to approximate $\sin(0.5)$, and bound the error.
Solution: The third-degree Maclaurin polynomial for $\sin x$ is:
$$P_3(x) = x - \frac{x^3}{3!} = x - \frac{x^3}{6}$$So $P_3(0.5) = 0.5 - \frac{(0.5)^3}{6} = 0.5 - \frac{0.125}{6} = 0.5 - 0.02083\overline{3} \approx 0.47917$.
For the Lagrange error bound with $n = 3$, we need $|f^{(4)}(c)| \leq M$ for $c$ between $0$ and $0.5$. The fourth derivative of $\sin x$ is $\sin x$, and $|\sin c| \leq 1$, so $M = 1$.
$$|R_3(0.5)| \leq \frac{1 \cdot |0.5|^4}{4!} = \frac{0.0625}{24} \approx 0.0026$$Our approximation $\sin(0.5) \approx 0.47917$ has an error of at most $0.0026$. (The actual value is $\sin(0.5) \approx 0.47943$.)
Approximate $\cos(1)$ using the first three nonzero terms of its Maclaurin series, and bound the error.
Solution: The Maclaurin series for $\cos x$ is $1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$
First three nonzero terms at $x = 1$:
$$1 - \frac{1}{2} + \frac{1}{24} = \frac{24 - 12 + 1}{24} = \frac{13}{24} \approx 0.54167$$This is an alternating series with decreasing terms. By the Alternating Series Error Bound, the error is at most the first omitted term:
$$|S - S_3| \leq \frac{1^6}{6!} = \frac{1}{720} \approx 0.00139$$So $\cos(1) \approx 0.54167$ with an error of at most $0.00139$. (The actual value is $\cos(1) \approx 0.54030$.)
10.8 Practice Problems
Test your understanding with these 12 problems covering all topics from this chapter. Click "Show Solution" to check your work.
Determine whether the sequence $a_n = \frac{(-1)^n n}{n+1}$ converges or diverges.
Show Solution
Consider the subsequences for even and odd $n$:
For even $n$: $a_n = \frac{n}{n+1} \to 1$.
For odd $n$: $a_n = \frac{-n}{n+1} \to -1$.
Since the subsequences approach different limits, the sequence diverges.
Find the sum of $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n 5}{3^n}$.
Show Solution
Rewrite: $\sum_{n=0}^{\infty} 5 \left(-\frac{1}{3}\right)^n$. This is geometric with $a = 5$ and $r = -\frac{1}{3}$.
Since $|r| = \frac{1}{3} < 1$, the series converges:
$$S = \frac{5}{1-(-1/3)} = \frac{5}{4/3} = \frac{15}{4}$$Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{2^n}{2^n + 1}$ converges or diverges.
Show Solution
$\lim_{n \to \infty} \frac{2^n}{2^n + 1} = \lim_{n \to \infty} \frac{1}{1 + 2^{-n}} = 1 \neq 0$.
The series diverges by the Divergence Test.
Determine whether $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n \ln n}$ converges or diverges.
Show Solution
Let $f(x) = \frac{1}{x \ln x}$, which is positive, continuous, and decreasing on $[2, \infty)$.
$$\int_2^{\infty} \frac{1}{x \ln x}\,dx = \lim_{b \to \infty} [\ln(\ln x)]_2^b = \lim_{b \to \infty} (\ln(\ln b) - \ln(\ln 2)) = \infty$$The integral diverges, so the series diverges by the Integral Test.
Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{\sin^2 n}{n^2}$ converges or diverges.
Show Solution
Since $0 \leq \sin^2 n \leq 1$, we have $0 \leq \frac{\sin^2 n}{n^2} \leq \frac{1}{n^2}$.
$\sum \frac{1}{n^2}$ converges (p-series, $p = 2$), so $\sum \frac{\sin^2 n}{n^2}$ converges by the Direct Comparison Test.
Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{n^3}{3^n}$ converges or diverges.
Show Solution
Apply the Ratio Test:
$$L = \lim_{n \to \infty} \frac{(n+1)^3/3^{n+1}}{n^3/3^n} = \lim_{n \to \infty} \frac{(n+1)^3}{3n^3} = \frac{1}{3} \lim_{n \to \infty} \left(\frac{n+1}{n}\right)^3 = \frac{1}{3}$$Since $L = \frac{1}{3} < 1$, the series converges by the Ratio Test.
Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt{n}}$ converges absolutely, conditionally, or diverges.
Show Solution
Absolute convergence: $\sum \frac{1}{\sqrt{n}} = \sum \frac{1}{n^{1/2}}$ is a p-series with $p = \frac{1}{2} \leq 1$, so it diverges. Not absolutely convergent.
Conditional convergence: The series is alternating with $b_n = \frac{1}{\sqrt{n}}$. Since $b_n$ is decreasing and $b_n \to 0$, the series converges by the Alternating Series Test.
The series converges conditionally.
Find the interval of convergence for $\displaystyle\sum_{n=1}^{\infty} \frac{(x+2)^n}{n \cdot 3^n}$.
Show Solution
Apply the Ratio Test:
$$L = \lim_{n \to \infty} \left|\frac{(x+2)^{n+1}}{(n+1) \cdot 3^{n+1}} \cdot \frac{n \cdot 3^n}{(x+2)^n}\right| = |x+2| \cdot \frac{1}{3} \cdot \lim_{n \to \infty} \frac{n}{n+1} = \frac{|x+2|}{3}$$Converges when $\frac{|x+2|}{3} < 1$, so $|x+2| < 3$. Thus $R = 3$ and center $a = -2$: $-5 < x < 1$.
At $x = 1$: $\sum \frac{3^n}{n \cdot 3^n} = \sum \frac{1}{n}$ diverges (harmonic).
At $x = -5$: $\sum \frac{(-3)^n}{n \cdot 3^n} = \sum \frac{(-1)^n}{n}$ converges (alternating harmonic).
Interval of convergence: $[-5, 1)$.
Find the first four nonzero terms of the Taylor series for $f(x) = \sqrt{x}$ centered at $a = 4$.
Show Solution
Compute derivatives at $a = 4$:
- $f(x) = x^{1/2} \Rightarrow f(4) = 2$
- $f'(x) = \frac{1}{2}x^{-1/2} \Rightarrow f'(4) = \frac{1}{4}$
- $f''(x) = -\frac{1}{4}x^{-3/2} \Rightarrow f''(4) = -\frac{1}{32}$
- $f'''(x) = \frac{3}{8}x^{-5/2} \Rightarrow f'''(4) = \frac{3}{256}$
Find the Maclaurin series for $f(x) = x^2 \cos(x^2)$ through the $x^{10}$ term.
Show Solution
Start with $\cos u = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \cdots$. Substitute $u = x^2$:
$$\cos(x^2) = 1 - \frac{x^4}{2} + \frac{x^8}{24} - \cdots$$Multiply by $x^2$:
$$x^2 \cos(x^2) = x^2 - \frac{x^6}{2} + \frac{x^{10}}{24} - \cdots$$Use the second-degree Maclaurin polynomial for $e^x$ to approximate $e^{0.1}$, and find the maximum error using the Lagrange error bound.
Show Solution
$P_2(x) = 1 + x + \frac{x^2}{2}$, so $P_2(0.1) = 1 + 0.1 + 0.005 = 1.105$.
For the error bound: $f^{(3)}(x) = e^x$. On $[0, 0.1]$, the maximum is $e^{0.1} < e < 3$, so $M = 3$.
$$|R_2(0.1)| \leq \frac{3 \cdot (0.1)^3}{3!} = \frac{3 \cdot 0.001}{6} = 0.0005$$So $e^{0.1} \approx 1.105$ with error at most $0.0005$. (Actual: $e^{0.1} \approx 1.10517$.)
How many terms of the series $\displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}$ are needed to approximate $\sin(1)$ with an error less than $0.001$?
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This is the Maclaurin series for $\sin x$ evaluated at $x = 1$: $\sin(1) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}$.
By the Alternating Series Error Bound, we need $\frac{1}{(2n+3)!} < 0.001$, where the error is bounded by the first omitted term.
- $n=0$: first omitted term is $\frac{1}{3!} = \frac{1}{6} \approx 0.167$ (too large)
- $n=1$: first omitted term is $\frac{1}{5!} = \frac{1}{120} \approx 0.0083$ (too large)
- $n=2$: first omitted term is $\frac{1}{7!} = \frac{1}{5040} \approx 0.000198 < 0.001$ ✓
We need 3 terms ($n = 0, 1, 2$): $\sin(1) \approx 1 - \frac{1}{6} + \frac{1}{120} = \frac{101}{120} \approx 0.84167$.