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AP Calculus BC : Series of Constants

Study concepts, example questions & explanations for AP Calculus BC

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2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Series And Functions

Consider: $\sum_{n=0}^{\infty} \left (\frac{\sqrt 2}{2}\right )^n$ . Will the series converge or diverge? If converges, where does this coverge to?

Possible Answers:

Diverge

$\sqrt{2}-2$

$2-\sqrt{2}$

$2+\sqrt2$

$\sqrt{2}$

Correct answer:

$2+\sqrt2$

Explanation:

This is a geometric series. Use the following formula, where is the first term of the series, and is the ratio that must be less than 1. If is greater than 1, the series diverges.

$\frac{a}{1-r} = \frac{1}{1-\frac{\sqrt2}{2}}= \frac{1}{\frac{2-\sqrt2}{2}} = \frac{2}{2-\sqrt2}$

Rationalize the denominator.

$\frac{2}{2-\sqrt2} \cdot \frac{2+\sqrt2}{2+\sqrt2} = \frac{2(2+\sqrt2)}{4-2}= 2+\sqrt2$

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Example Question #1 : Series And Functions

Consider the following summation: $1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...$ . Does this converge or diverge? If it converges, where does it approach?

Possible Answers:

Diverges

$\frac{7}{4}$

1.35

$\frac{4}{3}$

1.33

Correct answer:

$\frac{4}{3}$

Explanation:

The problem can be reconverted using a summation symbol, and it can be seen that this is geometric.

$1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...=\sum_{n=0}^{\infty} \left ( \frac{1}{4} \right )^n$

Since the ratio is less than 1, this series will converge. The formula for geometric series is:

$\frac{a}{1-r}$

where is the first term, and is the common ratio. Substitute these values and solve.

$\frac{a}{1-r} = \frac{1}{1-\frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$

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Example Question #3 : Geometric Series

A worm crawls up a wall during the day and slides down slowly during the night. The first day the worm crawls one meter up the wall. The first night the worm slides down a third of a meter. The second day the worm regains one third of the lost progress and slides down one third of that distance regained on the second night. This pattern of motion continues...

Which of the following is a geometric sum representing the distance the worm has travelled after 12-hour periods of motion? (Assuming day and night are both 12 hour periods).

Possible Answers:

$\sum_{k=0}^{n-1} (\frac{-1}{3})^{k}$

$\sum_{k=0}^{\infty} (\frac{-1}{3})^{k}$

$\sum_{k=0}^{n} (\frac{1}{3})^{k}$

$1+ \sum_{k=0}^{n} (\frac{-1}{3})^{k}$

$1+\sum_{k=1}^{n} (\frac{-1}{3})^{k}$

Correct answer:

$\sum_{k=0}^{n-1} (\frac{-1}{3})^{k}$

Explanation:

The sum must be alternating, and after one period you should have the worm at 1m. After two periods, the worm should be at 2/3m. There is only one sum for which that is true.

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Example Question #3 : Concepts Of Convergence And Divergence

Determine whether the following series converges or diverges. If it converges, what does it converge to?

$\sum_{n=0}^{\infty}(2)^{-2n}$

Possible Answers:

$\frac{3}{4}$

Diverges

$\frac{4}{3}$

Correct answer:

$\frac{4}{3}$

Explanation:

First, we reduce the series into a simpler form.

$\sum_{n=0}^{\infty}(2)^{-2n}= \sum_{n=0}^{\infty}(\frac{1}{2})^{2n}= \sum_{n=0}^{\infty}(\frac{1}{4})^{n}$

We know this series converges because

$\lim_{n \rightarrow \infty}(\frac{1}{4})^{n}=0$

By the Geometric Series Theorem, the sum of this series is given by

$\sum_{n=0}^{\infty}(\frac{1}{4})^{n}= \frac{(\frac{1}{4})^0}{1-\frac{1}{4}}= \frac{4}{3}$

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Example Question #1 : Series Of Constants

Calculate the sum of a geometric series with the following values: n=15 , $a_{1}=4$ , $r=\frac{1}{3}$ . Round the answer to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

This is a geometric series.

The sum of a geometric series can be calculated with the following formula,

$S_{n} = \frac{a_{1}*(1-r^n)}{1-r}$ , where n is the number of terms to sum up, r is the common ratio, and $a_{1}$ is the value of the first term.

For this question, we are given all of the information we need.

Solution:

$S_{15}=\frac{a_{1}*(1-r^n)}{1-r}$

$S_{15}=\frac{4*(1-(\frac{1}{3})^{15})}{1-\frac{1}{3}}$

$S_{15}=\frac{4*3(1-\frac{1}{3^{15}})}{2}$

$S_{15}=6*(1-\frac{1}{3^{15}})$

$S_{15}=5.999...$

Rounding, $S_{15}=6$

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Example Question #2 : Series Of Constants

Calculate the sum, rounded to the nearest integer, of the first 16 terms of the following geometric series: 6+12.6+26.46+116.6886+245.04606...

Possible Answers:

780295

780310

780305

780300

Correct answer:

780305

Explanation:

This is a geometric series.

The sum of a geometric series can be calculated with the following formula,

$S_{n} = \frac{a_{1}*(1-r^n)}{1-r}$ , where n is the number of terms to sum up, r is the common ratio, and $a_{1}$ is the value of the first term.

We have $a_{1}$ and n and we just need to find r before calculating the sum.

Solution:

$a_{1}=6$

$r=\frac{a_{2}}{a_{1}}$

$r=\frac{12.6}{6} = 2.1$

$S_{16}=\frac{a_{1}*(1-r^n)}{1-r}$

$S_{16}=\frac{6*(1-2.1^{16})}{1-2.1}$

$S_{16}=780305$

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Example Question #4 : Geometric Series

Calculate the sum of a geometric series with the following values:

n=10 , $a_{1}=11$ , $r=\frac{2}{20}$ ,

rounded to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

This is a geometric series.

The sum of a geometric series can be calculated with the following formula,

$S_{n} = \frac{a_{1}*(1-r^n)}{1-r}$ , where n is the number of terms to sum up, r is the common ratio, and $a_{1}$ is the value of the first term.

For this question, we are given all of the information we need.

Solution:

$S_{10}=\frac{a_{1}*(1-r^n)}{1-r}$

$S_{10}=\frac{11*(1-(\frac{2}{20})^{10})}{1-\frac{2}{20}}$

$S_{10}=\frac{11*(1-(\frac{1}{10})^{10}}{\frac{9}{10}} = \frac{11*10*(1-\frac{1}{10^{10}})}{9}$

$S_{10}=\frac{110}{9}(1-\frac{1}{10^{10}})$

$S_{10}=12.2222...$

Rounding, $S_{10}=12$

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Example Question #2 : Series Of Constants

Determine whether the following series converges or diverges:

$\sum_{n=0}^{\infty }(-1)^n(\frac{5}{n})$

Possible Answers:

The series (absolutely) converges

The series conditionally converges

The series diverges

The series may (absolutely) converge, diverge, or conditionally converge

Correct answer:

The series (absolutely) converges

Explanation:

Given just the harmonic series, we would state that the series diverges. However, we are given the alternating harmonic series. To determine whether this series will converge or diverge, we must use the Alternating Series test.

The test states that for a given series $\sum_{n=0}^{\infty }a_{n}$ where $a_{n}=(-1)^n(b_{n})$ or $a_{n}=(-1)^{n+1}(b_{n})$ where $b_{n}\geq0$ for all n, if $\lim_{n\rightarrow \infty }b_{n}=0$ and $b_{n}$ is a decreasing sequence, then $\sum_{n=0}^{\infty }a_{n}$ is convergent.

First, we must evaluate the limit of $b_{n}$ as n approaches infinity:

$\lim_{n\rightarrow \infty }\frac{5}{n}=5\lim_{n\rightarrow \infty }\frac{n}{n}(\frac{n^-1}{1})=0$

The limit equals zero because the numerator of the fraction equals zero as n approaches infinity.

Next, we must determine if $b_{n}$ is a decreasing sequence. $\frac{1}{n}< \frac{1}{n+1}$ , thus the sequence is decreasing.

Because both parts of the test passed, the series is (absolutely) convergent.

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Example Question #1 : Alternating Series

Determine whether

$\small \sum_{n=1}^{\infty} (-1)^n \sin\frac{1}{n}$

converges or diverges, and explain why.

Possible Answers:

Divergent, by the test for divergence.

Convergent, by the alternating series test.

More tests are needed.

Divergent, by the comparison test.

Convergent, by the $\small p$ -series test.

Correct answer:

Convergent, by the alternating series test.

Explanation:

We can use the alternating series test to show that

$\small \sum_{n=1}^{\infty} (-1)^n \sin\frac{1}{n}$

converges.

We must have $\small \sin \frac{1}{n}\geq 0$ for $\small n\geq 1$ in order to use this test. This is easy to see because $\small \frac{1}{n}$ is in $\small \small \left[0,\frac{\pi}{2}\right]$ for all $\small n\geq 1$ (the values of this sequence are $\small 1,1/2,1/3,1/4,...$ ), and sine is always nonzero whenever sine's argument is in $\small \small \left[0,\frac{\pi}{2}\right]$ .

Now we must show that

1. $\small \small \lim_{n\to\infty} \sin \frac{1}{n}=0$

2. $\small \sin \frac{1}{n}$ is a decreasing sequence.

The limit

$\small \lim_{n\to\infty}\frac{1}{n}=0$

implies that

$\small \small \small \lim_{n\to\infty} \sin \frac{1}{n}=\sin 0=0$

so the first condition is satisfied.

We can show that $\small \small \small \sin \frac{1}{n}$ is decreasing by taking its derivative and showing that it is less than $\small 0$ for $\small n\geq 1$ :

$\small \small \small \small \small \frac{d}{dn}\sin \frac{1}{n}=-\frac{1}{n^2}\cos\frac{1}{n}$

The derivative is less than $\small 0$ , because $\small -\frac{1}{n^2}$ is always less than $\small 0$ , and that $\small \cos \frac{1}{n}$ is positive for $\small n\geq 1$ , using a similar argument we used to prove that $\small \small \sin \frac{1}{n}\geq 0$ for $\small n\geq 1$ . Since the derivative is less than $\small 0$ , $\small \small \small \sin \frac{1}{n}$ is a decreasing sequence. Now we have shown that the two conditions are satisfied, so we have proven that

$\small \sum_{n=1}^{\infty} (-1)^n \sin\frac{1}{n}$

converges, by the alternating series test.

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Example Question #1 : Alternating Series With Error Bound

For the series: $\sum_{n=0}^{\infty} (-1)^n (\frac{n^n}{8^n})$ , determine if the series converge or diverge. If it diverges, choose the best reason.

Possible Answers:

$\textup{Converge, since all Alternating Series Test rules are satisfied.}$

$\textup{Diverges, since the terms are not progressively decreasing}.$

$\textup{Converge, since}\lim_{n \to \infty}x_n\neq 0.$

$\textup{Diverges, by the Geometric Series Test.}$

$\textup{Diverge, since}\lim_{n \to \infty}x_n\neq 0.$

Correct answer:

$\textup{Diverge, since}\lim_{n \to \infty}x_n\neq 0.$

Explanation:

The series given is an alternating series.

Write the three rules that are used to satisfy convergence in an alternating series test.

For $\sum_{n=1}^{\infty} (-1)^{n+1} x_n= x_1-x_2+x_3-x_4+...$ :

$\textup{1. All the } x_n \textup{ terms are positive.}$

$\textup{2. The terms must be decreasing: } x_n \geq x_{n+1}$

$\textup{3. The limit }\lim_{n \to \infty}x_n=0.$

$\sum_{n=0}^{\infty} (-1)^n (\frac{n^n}{8^n})=\sum_{n=0}^{\infty} (-1)^n (\frac{n}{8})^n$

The first and second conditions are satisfied since the terms are positive and are decreasing after each term.

However, the third condition is not valid since $\lim_{n \to \infty}x_n\neq 0$ and instead approaches infinity.

The correct answer is:

$\textup{Diverge, since}\lim_{n \to \infty}x_n\neq 0.$

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AP Calculus BC : Series of Constants

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #1 : Series And Functions

Example Question #1 : Series And Functions

Example Question #3 : Geometric Series

Example Question #3 : Concepts Of Convergence And Divergence

Example Question #1 : Series Of Constants

Example Question #2 : Series Of Constants

Example Question #4 : Geometric Series

Example Question #2 : Series Of Constants

Example Question #1 : Alternating Series

Example Question #1 : Alternating Series With Error Bound

All AP Calculus BC Resources

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