Unit 9BC only

Sequences & Series

When a series converges and how to tell, Taylor and Maclaurin series, intervals of convergence, and error bounds.

Topics in this unit

Sequences · Series & Geometric Series · Divergence & Integral Test · Comparison Tests · Alternating Series · Ratio & Root Tests · Absolute & Conditional Convergence · Convergence Strategy · Determining Convergence · Taylor & Maclaurin Polynomials · Error Bounds · Radius & Interval of Convergence · Taylor & Maclaurin Series · Power Series Operations

Sequences

Question 1

BCno calculatormedium

Which of the following sequences converge? I.

a_{n} = \frac{n ^{2}}{2 n ^{2} + 1}

II.

a_{n} = \frac{e ^{n}}{n}

III.

a_{n} = \frac{e ^{n}}{e ^{n} + 1}

Show the answer

The answer is A.

\frac{n ^{2}}{2 n ^{2} + 1} \to \frac{1}{2}

(converges). II:

\frac{e ^{n}}{n} \to \infty

(diverges). III:

\frac{e ^{n}}{e ^{n} + 1} = \frac{1}{1 + e ^{- n}} \to 1

(converges). So I and III only.

Why not B:III also converges: $\frac{e ^{n}}{e ^{n} + 1} = \frac{1}{1 + e ^{- n}} \to 1$ as $n \to \infty$ .
Why not C:II diverges: $\frac{e ^{n}}{n} \to \infty$ (exponential beats linear). I converges to $\frac{1}{2}$ .
Why not D:II diverges to $\infty$ . Only I ( $\to \frac{1}{2}$ ) and III ( $\to 1$ ) converge.

Series & Geometric Series

Question 2

BCno calculatormedium

For what values of

x

does the series

n = 1 \sum \infty (\frac{2}{x ^{2} + 1})^{n}

converge?

Show the answer

The answer is D.

It is geometric with ratio

r = \frac{2}{x ^{2} + 1} > 0

; it converges iff

r < 1

, i.e.

\frac{2}{x ^{2} + 1} < 1 \Rightarrow x^{2} + 1 > 2 \Rightarrow x^{2} > 1

, so

x < - 1

x > 1

Why not A:Solved the inequality backwards: $\frac{2}{x ^{2} + 1} < 1$ gives $x^{2} > 1$ , i.e. $x$ OUTSIDE $[- 1, 1]$ .
Why not B:Dropped the negative branch; $x^{2} > 1$ also allows $x < - 1$ .
Why not C:A geometric series needs strict $∣ r ∣ < 1$ ; at $x = \pm 1$ , $r = 1$ and the series diverges.

Comparison Tests

Question 3

BCno calculatorhard

Suppose

0 \leq a_{n} \leq b_{n}

for all

n

. Which of the following must be true?

Show the answer

The answer is C.

By the Direct Comparison Test, if the smaller nonnegative series

\sum a_{n}

diverges, the larger series

\sum b_{n}

must diverge too. The test runs 'small diverges

\Rightarrow

big diverges' and 'big converges

\Rightarrow

small converges.'

Why not A:Divergence of the larger series tells us nothing about the smaller one; $a_{n} = \frac{1}{n ^{2}} \leq b_{n} = \frac{1}{n}$ with $\sum b_{n}$ divergent yet $\sum a_{n}$ convergent.
Why not B:Convergence of the smaller series does not force the larger one to converge; $a_{n} = \frac{1}{n ^{2}} \leq b_{n} = \frac{1}{n}$ has $\sum a_{n}$ convergent but $\sum b_{n}$ divergent.
Why not D:If $\sum b_{n}$ converges, comparison forces the smaller $\sum a_{n}$ to converge, not diverge.

Alternating Series

Question 4

BCno calculatormedium

For

n = 1 \sum \infty \frac{( - 1 ) ^{n}}{n}

, which statement is correct?

Show the answer

The answer is D.

b_{n} = \frac{1}{n}

is positive, decreasing, and

\to 0

, so the alternating series test confirms convergence (it is conditional, since

\sum \frac{1}{n}

diverges).

Why not A: $\frac{1}{n} \to 0$ and is decreasing, so the alternating series test gives convergence.
Why not B: $\sum \frac{1}{n}$ is a divergent $p$ -series ( $p = \frac{1}{2}$ ), so convergence is conditional, not absolute.
Why not C:The terms $\frac{1}{n}$ are positive, decreasing, and $\to 0$ , so the test applies and gives convergence.

Ratio & Root Tests

Question 5

BCno calculatormedium

Suppose

n \to \infty lim \frac{a _{n + 1}}{a _{n}} = 1

. Which of the following must be true?

Show the answer

The answer is A.

When the ratio limit equals

1

, the Ratio Test yields no conclusion. Both

\sum \frac{1}{n}

(diverges) and

\sum \frac{1}{n ^{2}}

(converges) have ratio limit

1

, so neither convergence nor divergence is forced.

Why not B:A ratio limit of $1$ does not imply divergence; $\sum \frac{1}{n ^{2}}$ has ratio limit $1$ and converges.
Why not C:A ratio limit of $1$ does not imply convergence; $\sum \frac{1}{n}$ has ratio limit $1$ and diverges.
Why not D:The ratio of consecutive terms approaching $1$ says nothing about the size of $a_{n}$ ; e.g. $a_{n} = \frac{1}{n} \to 0$ , not $1$ .

Question 6

BCno calculatorhard

n = 1 \sum \infty \frac{n !}{n ^{n}}

converges by:

Show the answer

The answer is A.

Ratio Test:

\frac{a _{n + 1}}{a _{n}} = \frac{( n + 1 )!}{( n + 1 ) ^{n + 1}} \cdot \frac{n ^{n}}{n !} = \frac{n ^{n}}{( n + 1 ) ^{n}} = (\frac{n}{n + 1})^{n} \to \frac{1}{e} < 1

. Converges.

Why not B:The Integral Test requires integrating $n! / n^{n}$ , which is extremely difficult. The Ratio Test is the natural choice for factorials.
Why not C:Finding a comparison series for $n! / n^{n}$ is tricky. The Ratio Test handles factorials cleanly.
Why not D:The series converges since the ratio limit $1/ e < 1$ .

Error Bounds

Question 7

BCno calculatorhard

The number

e^{- 1} = n = 0 \sum \infty \frac{( - 1 ) ^{n}}{n !}

is approximated by a partial sum. What is the fewest terms needed so the alternating-series error is less than

\frac{1}{100}

Show the answer

The answer is C.

Error

\leq

first omitted term

\frac{1}{m !}

. Need

\frac{1}{m !} < \frac{1}{100}

\frac{1}{4 !} = \frac{1}{24}

(too big),

\frac{1}{5 !} = \frac{1}{120} < \frac{1}{100}

. So include

n = 0, \dots, 4

5

terms.

Why not A:With $4$ terms ( $n = 0..3$ ) the first omitted is $\frac{1}{4 !} = \frac{1}{24} > \frac{1}{100}$ ; you need the next term too.
Why not B: $5$ terms ( $n = 0..4$ ) already give first omitted $\frac{1}{5 !} = \frac{1}{120} < \frac{1}{100}$ , so $6$ is more than needed.
Why not D:Factorials shrink fast: $\frac{1}{5 !} = \frac{1}{120} < \frac{1}{100}$ , so only $5$ terms are required.

Radius & Interval of Convergence

Question 8

BCno calculatorhard

What is the interval of convergence of

n = 1 \sum \infty \frac{( 3 x - 2 ) ^{n}}{n}

Show the answer

The answer is C.

The ratio test gives

∣ 3 x - 2 ∣ < 1

, so

- 1 < 3 x - 2 < 1

and

\frac{1}{3} < x < 1

. At

x = 1

3 x - 2 = 1

and the series is harmonic, which diverges. At

x = \frac{1}{3}

3 x - 2 = - 1

and the series is alternating harmonic, which converges. The interval is

\frac{1}{3} \leq x < 1

Why not A:This solves $∣ x - 2 ∣ < 1$ , forgetting the coefficient $3$ on $x$ . Divide by $3$ : the center is $\frac{2}{3}$ and the radius is $\frac{1}{3}$ .
Why not B:This leaves out $x = \frac{1}{3}$ . There the series is $\sum \frac{( - 1 ) ^{n}}{n}$ , which converges.
Why not D:This swaps the endpoints. At $x = 1$ every term is $\frac{1}{n}$ , so the series diverges there.

Taylor & Maclaurin Series

Question 9

BCno calculatorhard

The Maclaurin series for a function

f

n = 0 \sum \infty \frac{n + 1}{2 ^{n}} x^{n}

. What is

f^{'''} (0)

Show the answer

The answer is D.

In a Maclaurin series

f (x) = \sum a_{n} x^{n}

, the coefficient is

a_{n} = \frac{f ^{(n)} ( 0 )}{n !}

, so

f^{(n)} (0) = n! a_{n}

. Here

a_{3} = \frac{3 + 1}{2 ^{3}} = \frac{1}{2}

, so

f^{'''} (0) = 3! \cdot \frac{1}{2} = 6 \cdot \frac{1}{2} = 3

Why not A: $\frac{1}{2} = a_{3}$ is the coefficient itself; the derivative is $f^{'''} (0) = 3! a_{3}$ , not $a_{3}$ .
Why not B: $\frac{3}{2} = 3 \cdot a_{3}$ multiplies by $n = 3$ ; the correct factor is $n! = 3! = 6$ , giving $f^{'''} (0) = 3$ .
Why not C: $6 = 3! \cdot 1$ assumes $a_{3} = 1$ , but $a_{3} = \frac{3 + 1}{2 ^{3}} = \frac{1}{2}$ , so $f^{'''} (0) = 3$ .

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