Unit 10 · 15 to 20% of the exam

Unit 10 practice test: Infinite Sequences and Series

twelve questions, every option explained

12 exam-style questions from the unit, free, with an explanation for every option. Most have four answer choices, as on the real exam, and a few ask you to type a value or an expression; sets share one graph or table. Covers: Convergent and Divergent Series; Geometric Series; The nth Term Test for Divergence; The Integral Test; Harmonic Series and p-Series; Comparison Tests; The Alternating Series Test; The Ratio Test; Absolute and Conditional Convergence; The Alternating Series Error Bound; Taylor Polynomials; The Lagrange Error Bound; Radius and Interval of Convergence; Taylor and Maclaurin Series; Representing Functions as Power Series.

Start at question 1

1. Does $\sum_{n=0}^{\infty} \frac{3^n}{2^n}$ converge?

Yes: to $\frac{1}{1 - 3/2} = -2$
Yes: it is geometric
No: its ratio $\frac{2}{3}$ is less than 1
No: it is geometric with $r = \frac{3}{2}$ , and $|r| \ge 1$

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$\frac{3^n}{2^n} = \left(\frac{3}{2}\right)^n$ : geometric with $r = \frac{3}{2}$ . Since $|r| \ge 1$ , the series diverges.

Yes: to $\frac{1}{1 - 3/2} = -2$ : The formula $\frac{a}{1 - r}$ holds only when $|r| < 1$ . A sum of positive terms cannot be $-2$ .
Yes: it is geometric: Being geometric is not enough: the ratio $\frac{3}{2}$ is too big.
No: its ratio $\frac{2}{3}$ is less than 1: The ratio is $\frac{3^{n+1}/2^{n+1}}{3^n/2^n}$ $= \frac{3}{2}$ , not $\frac{2}{3}$ .
Correct: No: it is geometric with $r = \frac{3}{2}$ , and $|r| \ge 1$ : Right: the terms grow, and the partial sums grow without bound.

2. What does the integral test say about $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ ?

It converges, since $\frac{1}{\sqrt{n}} \to 0$ as $n$ grows
It converges: $\int_1^{\infty} x^{-1/2}\,dx$ $= 2$
Nothing: $\frac{1}{\sqrt{x}}$ is not decreasing
It diverges, since $\int_1^{\infty} x^{-1/2}\,dx$ diverges

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$f(x) = \frac{1}{\sqrt{x}}$ is positive, continuous and decreasing on $[1, \infty)$ , and $\lim_{b \to \infty} \left(2\sqrt{b} - 2\right)$ $= \infty$ , so the series diverges by the integral test.

It converges, since $\frac{1}{\sqrt{n}} \to 0$ as $n$ grows: Terms approaching 0 never prove convergence.
It converges: $\int_1^{\infty} x^{-1/2}\,dx$ $= 2$ : $\int_1^{b} x^{-1/2}\,dx$ $= 2\sqrt{b} - 2$ , and $\sqrt{b}$ grows without bound as $b \to \infty$ .
Nothing: $\frac{1}{\sqrt{x}}$ is not decreasing: $\frac{1}{\sqrt{x}}$ is positive, continuous and decreasing for $x \ge 1$ .
Correct: It diverges, since $\int_1^{\infty} x^{-1/2}\,dx$ diverges: Right: the integral diverges, so the series does.

3. What does the direct comparison test say about $\sum_{n=1}^{\infty} \frac{1}{n^2 + 5}$ ?

It diverges, since $\frac{1}{n^2 + 5} < \frac{1}{n}$
It converges, since $0 < \frac{1}{n^2 + 5} < \frac{1}{n^2}$
It converges, since $\frac{1}{n^2 + 5} > \frac{1}{n^3}$
It diverges, since $n^2 + 5 > n^2$

Show the answer

For every $n$ , $0 < \frac{1}{n^2 + 5} < \frac{1}{n^2}$ , and $\sum \frac{1}{n^2}$ converges ( $p = 2 > 1$ ), so the series converges by the direct comparison test.

It diverges, since $\frac{1}{n^2 + 5} < \frac{1}{n}$ : Smaller than a divergent series decides nothing.
Correct: It converges, since $0 < \frac{1}{n^2 + 5} < \frac{1}{n^2}$ : Right: smaller than a convergent $p$ -series with $p = 2$ .
It converges, since $\frac{1}{n^2 + 5} > \frac{1}{n^3}$ : Larger than a convergent series decides nothing.
It diverges, since $n^2 + 5 > n^2$ : A larger denominator makes the terms smaller than $\frac{1}{n^2}$ , which points to convergence.

Table

The sizes of the terms of an alternating series

$a_n = \frac{n}{n^2 + 3}$ for the series $\sum_{n=1}^{\infty} (-1)^n \frac{n}{n^2 + 3}$
$n$	$1$	$2$	$3$	$4$	$5$
$a_n$	$\frac{1}{4}$	$\frac{2}{7}$	$\frac{3}{12}$	$\frac{4}{19}$	$\frac{5}{28}$

4. What does the table show about the sizes $a_n$ ?

The sizes decrease from $n = 1$ on, as the test needs
$a_2 > a_1$ , and from $n = 2$ on the sizes decrease
The sizes increase from $n = 1$ to the last entry
The sizes approach $\frac{1}{4}$ as $n$ grows

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$a_1 = 0.25$ and $a_2 \approx 0.286$ , so the sizes rise once; then $a_3 = 0.25$ , $a_4 \approx 0.211$ , $a_5 \approx 0.179$ keep falling.

The sizes decrease from $n = 1$ on, as the test needs: $\frac{2}{7} \approx 0.286$ is larger than $\frac{1}{4} = 0.25$ .
Correct: $a_2 > a_1$ , and from $n = 2$ on the sizes decrease: Right: $\frac{2}{7} > \frac{1}{4}$ , then the values fall.
The sizes increase from $n = 1$ to the last entry: From $n = 2$ the sizes fall: $\frac{2}{7}, \frac{3}{12}, \frac{4}{19}, \frac{5}{28}$ .
The sizes approach $\frac{1}{4}$ as $n$ grows: The sizes behave like $\frac{n}{n^2} = \frac{1}{n}$ and approach 0.

5. Which argument shows that $a_n = \frac{n}{n^2 + 3}$ decreases for $n \ge 2$ ?

The table values fall from $n = 2$ to $n = 5$
The numerator $n$ grows more slowly than $n^2 + 3$ , so the terms approach 0
With $f(x) = \frac{x}{x^2 + 3}$ , $f'(x)$ $= \frac{3 - x^2}{(x^2 + 3)^2} < 0$ for $x > \sqrt{3}$
With $f(x) = \frac{x}{x^2 + 3}$ , $f'(x) = \frac{1}{2x} < 0$ for $x > 0$

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By the quotient rule, $f'(x)$ $= \frac{(x^2 + 3) - x(2x)}{(x^2 + 3)^2}$ $= \frac{3 - x^2}{(x^2 + 3)^2}$ , which is negative for $x > \sqrt{3} \approx 1.73$ . So $a_{n+1} < a_n$ for every $n \ge 2$ .

The table values fall from $n = 2$ to $n = 5$ : Five values cannot show what happens for every larger $n$ ; a derivative argument can.
The numerator $n$ grows more slowly than $n^2 + 3$ , so the terms approach 0: Approaching 0 is a different condition from decreasing; a sequence can approach 0 while going up and down.
Correct: With $f(x) = \frac{x}{x^2 + 3}$ , $f'(x)$ $= \frac{3 - x^2}{(x^2 + 3)^2} < 0$ for $x > \sqrt{3}$ : Right: a negative derivative past $\sqrt{3}$ covers every $n \ge 2$ .
With $f(x) = \frac{x}{x^2 + 3}$ , $f'(x) = \frac{1}{2x} < 0$ for $x > 0$ : That differentiates the top and bottom separately; use the quotient rule.

6. What does the ratio test conclude about $\sum_{n=1}^{\infty} \frac{(-1)^n n}{4^n}$ ?

It does not apply, since the terms change sign
It diverges: the ratio $\frac{a_{n+1}}{a_n}$ is negative
It converges: the limit is 4
It converges: $\left|\frac{a_{n+1}}{a_n}\right| \to \frac{1}{4}$

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$\left|\frac{a_{n+1}}{a_n}\right|$ $= \frac{n + 1}{4^{n+1}} \cdot \frac{4^n}{n}$ $= \frac{n + 1}{4n} \to \frac{1}{4} < 1$ , so the series converges by the ratio test.

It does not apply, since the terms change sign: The ratio test uses $\left|\frac{a_{n+1}}{a_n}\right|$ , so it works on series with signs.
It diverges: the ratio $\frac{a_{n+1}}{a_n}$ is negative: The test takes the absolute value of the ratio before the limit.
It converges: the limit is 4: That inverts the ratio; a limit of 4 would mean divergence.
Correct: It converges: $\left|\frac{a_{n+1}}{a_n}\right| \to \frac{1}{4}$ : Right: the absolute value removes the sign, and $\frac{1}{4} < 1$ .

7. For which values of $p$ is $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^p}$ conditionally convergent?

$0 < p \le 1$
$p > 1$
$p \le 0$
$p \ge 1$

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For $p > 0$ the sizes $\frac{1}{n^p}$ decrease to 0, so the series converges. The absolute values form a $p$ -series, which diverges for $p \le 1$ . So the convergence is conditional for $0 < p \le 1$ and absolute for $p > 1$ .

Correct: $0 < p \le 1$ : Right: then it converges by the alternating series test, while $\sum \frac{1}{n^p}$ diverges.
$p > 1$ : For $p > 1$ , $\sum \frac{1}{n^p}$ converges, so the convergence is absolute.
$p \le 0$ : For $p \le 0$ the sizes do not approach 0, and the series diverges.
$p \ge 1$ : For $p > 1$ the convergence is absolute; $p = 1$ is the one conditional case in that range.

8. For an alternating series that meets the conditions of the test, $S_{20} = 1.3527$ and $a_{21} = 0.01$ . What does the error bound tell you about the sum $S$ ?

$1.3427 \le S \le 1.3627$
$S = 1.3527 \pm 0.01$ exactly
$S = 1.3627$
$S \ge 1.3527$

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$|S - S_{20}| \le a_{21} = 0.01$ means $S_{20} - 0.01 \le S \le S_{20} + 0.01$ , that is $1.3427 \le S \le 1.3627$ .

Correct: $1.3427 \le S \le 1.3627$ : Right: $S$ is within $0.01$ of $S_{20}$ .
$S = 1.3527 \pm 0.01$ exactly: The bound gives a range for $S$ , not two exact values.
$S = 1.3627$ : The bound limits the error; it does not say the error equals $0.01$ .
$S \ge 1.3527$ : The sum may lie on either side of $S_{20}$ , depending on the sign of the 21st term.

9. Why is the fifth-degree Maclaurin polynomial for $\cos x$ the same as the fourth-degree one?

The fifth-degree term is too small to matter
$\cos x$ has no derivatives past the fourth
A polynomial for $\cos x$ stops at degree 4
$\frac{d^5}{dx^5}\cos x = -\sin x$ , which is 0 at $x = 0$

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The derivatives of $\cos x$ cycle: $\cos x,$ $-\sin x,$ $-\cos x,$ $\sin x,$ $\cos x,$ $-\sin x,$ $\ldots$ At 0, every odd-order derivative is 0, so the $x^5$ term is $0 \cdot x^5$ and $P_5 = P_4$ .

The fifth-degree term is too small to matter: The coefficient is exactly 0, not small.
$\cos x$ has no derivatives past the fourth: $\cos x$ has derivatives of every order; they cycle with period 4.
A polynomial for $\cos x$ stops at degree 4: It can have any even degree; the odd-degree coefficients are 0.
Correct: $\frac{d^5}{dx^5}\cos x = -\sin x$ , which is 0 at $x = 0$ : Right: the $x^5$ coefficient is 0.

10. $P_3(x) = x - \frac{x^3}{6}$ approximates $\sin x$ . Using $\left|\frac{d^4}{dx^4}\sin x\right| \le 1$ , which is the Lagrange error bound for $\left|\sin(0.2) - P_3(0.2)\right|$ ?

$\frac{(0.2)^3}{3!} = \frac{1}{750}$
$(0.2)^4 = \frac{1}{625}$ , with no factorial
$\frac{(0.2)^4}{4!} = \frac{1}{15000}$
$\frac{(0.2)^4}{3!} = \frac{1}{3750}$

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$\left|\sin(0.2) - P_3(0.2)\right| \le \frac{1}{4!}(0.2)^4$ $= \frac{0.0016}{24}$ $= \frac{1}{15000}$ .

$\frac{(0.2)^3}{3!} = \frac{1}{750}$ : The bound for $P_3$ uses the next derivative: the fourth power and $4!$ .
$(0.2)^4 = \frac{1}{625}$ , with no factorial: Divide by $4! = 24$ .
Correct: $\frac{(0.2)^4}{4!} = \frac{1}{15000}$ : Right: the fourth derivative, $4!$ , and the fourth power.
$\frac{(0.2)^4}{3!} = \frac{1}{3750}$ : The factorial matches the power: $4!$ with $(0.2)^4$ .

11. A power series $\sum a_n (x + 1)^n$ converges at $x = 3$ and diverges at $x = -7$ . What can you conclude about its radius of convergence $R$ ?

$R = 4$
$R = 6$
$4 \le R \le 6$
$R \ge 6$

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The center is $-1$ . Converging at $x = 3$ , a distance 4 from the center, means $R \ge 4$ . Diverging at $x = -7$ , a distance 6 away, means $R \le 6$ . So $4 \le R \le 6$ .

$R = 4$ : $R$ could be larger than 4; the series could converge farther out.
$R = 6$ : $R$ could be smaller than 6, down to 4.
Correct: $4 \le R \le 6$ : Right: converging 4 from the center needs $R \ge 4$ ; diverging 6 away needs $R \le 6$ .
$R \ge 6$ : The series diverges at distance 6 from the center, so $R$ is at most 6.

12. How is the Taylor polynomial $P_4(x)$ $= 1$ $+ x$ $+ \frac{x^2}{2}$ $+ \frac{x^3}{6}$ $+ \frac{x^4}{24}$ related to the Maclaurin series for $e^x$ ?

It equals the series for $x$ between $-1$ and $1$
It is a partial sum: the series' first five terms
It is the remainder left after the series
It is unrelated: the series has different coefficients

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The Maclaurin series for $e^x$ is $\sum \frac{x^n}{n!}$ , and $P_4$ is its partial sum through $n = 4$ : its first five terms.

It equals the series for $x$ between $-1$ and $1$ : The series has more nonzero terms; $P_4$ only approximates it.
Correct: It is a partial sum: the series' first five terms: Right: a Taylor polynomial is a partial sum of the Taylor series.
It is the remainder left after the series: $P_4$ is the beginning of the series, not what is left over.
It is unrelated: the series has different coefficients: The coefficients are the same, $\frac{1}{n!}$ ; the series keeps going.