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Calculus 2 : Calculus II

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #55 : Types Of Series

Determine if the following series is convergent or divergent using the alternating series test:

$\sum_{n=1}^{\infty} cos(n\pi)\frac{7(n^{11}+n^7+n^3)}{n^{20}}$

Possible Answers:

Convergent according to the alternating series test

Divergent according to the alternating series test

Divergent according to the ratio test

Inconclusive according to the alternating series test

Correct answer:

Convergent according to the alternating series test

Explanation:

This is an alternating series.

An alternating series can be identified because terms in the series will “alternate” between + and –, because of $cos(n\pi)$

Note: Alternating Series Test can only show convergence. It cannot show divergence.

If the following 2 tests are true, the alternating series converges.

$a_{n} = cos(n\pi) * b_{n}$

$\lim_{n \rightarrow \infty} b_{n} = 0$
{ $b_{n}$ } is a decreasing sequence, or in other words $b_{n=1}' < 0$

Solution:

$b_{n}=\frac{7(n^{11}+n^7+n^3)}{n^{20}}$

1. $\lim_{n \rightarrow \infty}b_{n}$

$\lim_{n \rightarrow \infty}\frac{7(n^{11}+n^7+n^3)}{n^{20}} =0$

2. $b_{n}'=(\frac{7(n^{11}+n^7+n^3)}{n^{20}})'$

$b_{n}'=7(\frac{n^{11}+n^7+n^3}{n^{20}})'$

$b_{n}'= 7(\frac{n^{20}(n^{11}+n^7+n^3)' - (n^{20})'(n^11+n^7+n^3)}{n^{40}}$

$b_{n}'=\frac{7((11n^{10}+7n^{6}+3n^2)*n^{20} - 20n^{19}(n^{11}+n^7+n^3))}{n^{40}}$ $b_{1}'=\frac{7((11*1^{10}+7*1^{6}+3*1^2)*1^{20} - 20*1^{19}(1^{11}+1^7+1^3))}{1^{40}}$ $b_{1}'= \frac{7(11+7+3-20(1+1+1)}{1}=\frac{7(21-60)}{1} = -39*7 < 0$

Since the 2 tests pass, this series is convergent.

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Example Question #61 : Types Of Series

Determine how many terms need to be added to approximate the following series within $\frac{1}{1000}$ :

$\sum_{n=1}^{\infty} (-1)^n\frac{10}{n!}$

Possible Answers:

Correct answer:

Explanation:

This is an alternating series test.

In order to find the terms necessary to approximate the series within $\frac{1}{1000}$ first see if the series is convergent using the alternating series test. If the series converges, find n such that $b_{n+1} < \frac{1}{1000}$

Step 1:

An alternating series can be identified because terms in the series will “alternate” between + and –, because of (-1)^n

Note: Alternating Series Test can only show convergence. It cannot show divergence.

If the following 2 tests are true, the alternating series converges.

$a_{n} = (-1)^n * b_{n}$

$\lim_{n \rightarrow \infty} b_{n} = 0$
{ $b_{n}$ } is a decreasing sequence, or in other words $b_{n=1}' < 0$

Solution:

$b_{n}=\frac{10}{n!}$

$1. \lim_{n \rightarrow \infty}b_{n}$

$\lim_{n \rightarrow \infty}\frac{10}{n!} =0$

2. { $b_{n}$ } is a decreasing sequence

$b_{1} = \frac{10}{1},b_{2}=\frac{10}{2}, b_{3} = \frac{10}{3*2}...$

Since the factorial is always increasing, the sequence is always decreasing.

Since the 2 tests pass, this series is convergent.

Step 2:

Plug in n values until $b_{n+1} < \frac{1}{1000}$

$b_{7} = \frac{1}{504} > \frac{1}{1000}$

$b_{8} = \frac{1}{4032} < \frac{1}{1000}$

so 7 terms are needed to approximate the sum within .001.

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Example Question #3021 : Calculus Ii

Determine how many terms need to be added to approximate the following series within $\frac{1}{1000}$ :

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

Possible Answers:

Correct answer:

Explanation:

This is an alternating series test.

Step 1:

An alternating series can be identified because terms in the series will “alternate” between + and –, because of (-1)^n

Note: Alternating Series Test can only show convergence. It cannot show divergence.

If the following 2 tests are true, the alternating series converges.

$a_{n} = (-1)^n * b_{n}$

$\lim_{n \rightarrow \infty} b_{n} = 0$
{ $b_{n}$ } is a decreasing sequence, or in other words $b_{n=1}' < 0$

Solution:

$b_{n}=\frac{4}{(n!)^2}$

1. $\lim_{n \rightarrow \infty}b_{n}$

$\lim_{n \rightarrow \infty}\frac{4}{(n!)^2} =0$

2. { $b_{n}$ } is a decreasing functon, since a factorial never decreases.

Since the 2 tests pass, this series is convergent.

Step 2:

Plug in n values until $b_{n+1} < \frac{1}{1000}$

$b_{4}=\frac{4}{(4!)^2}=\frac{1}{144} > \frac{1}{1000}$

$b_{5}=\frac{4}{(5!)^2}=\frac{1}{3600} < \frac{1}{1000}$

4 needs to be added to approximate the sum within $\frac{1}{1000}$ .

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Example Question #19 : Alternating Series

Determine how many terms need to be added to approximate the following series within $\frac{1}{1000}$ :

$\sum_{n=1}^{\infty} cos(n\pi)\frac{13}{n^5}$

Possible Answers:

Correct answer:

Explanation:

This is an alternating series test.

Step 1:

An alternating series can be identified because terms in the series will “alternate” between + and –, because of $cos(n\pi)$

Note: Alternating Series Test can only show convergence. It cannot show divergence.

If the following 2 tests are true, the alternating series converges.

$a_{n} = cos(n\pi) * b_{n}$

$\lim_{n \rightarrow \infty} b_{n} = 0$
{ $b_{n}$ } is a decreasing sequence, or in other words $b_{n=1}' < 0$

Solution:

$b_{n}=\frac{13}{n^5}$

1. $\lim_{n \rightarrow \infty}b_{n}$

$\lim_{n \rightarrow \infty}\frac{13}{n^5} =0$

2. $b_{n}'=(\frac{13}{n^5})'$

$b_{n}'=13(\frac{1}{n^5})' = -13(\frac{(n^5)'}{n^{10}})$

$b_{n}'=-13*5\frac{n^4}{n^{10}} = -65\frac{1}{n^6}$

$b_{1}'=-65\frac{1}{1^6} = -65 < 0$

Since the 2 tests pass, this series is convergent.

Step 2:

Plug in n values until $b_{n+1} < \frac{1}{1000}$

$b_{6}=\frac{13}{6^5} = \frac{13}{7776} > \frac{1}{1000}$

$b_{7}=\frac{13}{7^5} = \frac{13}{16807} < \frac{1}{1000}$

7 terms are needed to approximate the sum within .001

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Example Question #21 : Alternating Series

Determine how many terms need to be added to approximate the following series within

$\frac{1}{1000}: \sum_{n=1}^{\infty} cos(n\pi)\frac{50}{n!^3}$

Possible Answers:

Correct answer:

Explanation:

This is an alternating series test.

Step 1:

An alternating series can be identified because terms in the series will “alternate” between + and –, because of $cos(n\pi)$

Note: Alternating Series Test can only show convergence. It cannot show divergence.

If the following 2 tests are true, the alternating series converges.

$a_{n} = cos(n\pi) * b_{n}$

$\lim_{n \rightarrow \infty} b_{n} = 0$
{ $b_{n}$ } is a decreasing sequence, or in other words $b_{n=1}' < 0$

Solution:

$b_{n}=\frac{50}{n!^3}$

1. $\lim_{n \rightarrow \infty}b_{n}$

$\lim_{n \rightarrow \infty}\frac{50}{n!^3} =0$

2. {{b_{n}} is a decreasing sequence. A factorial always increases as n increases, so each term will decrease as n increases.

Since the 2 tests pass, this series is convergent.

Step 2:

Plug in n values until $b_{n+1} < \frac{1}{1000}$

$b_{4}=\frac{50}{4!^3} = \frac{25}{6912} > \frac{1}{1000}$

$b_{5}=\frac{13}{7^5} = \frac{1}{34560} < \frac{1}{1000}$

4 terms need to be added to approximate the sum within .001

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Example Question #22 : Alternating Series

Determine whether the series converges or diverges:

$\sum_{n=0}^{\infty }(-1)^n(\frac{2n}{n^3})$

Possible Answers:

The series is conditionally convergent

The series is divergent

The series may be absolutely convergent, conditionally convergent, or divergent

The series is (absolutely) convergent

Correct answer:

The series is (absolutely) convergent

Explanation:

To determine whether the given alternating series converges or diverges, we must perform the Alternating Series test, which states that for a given series

$\sum_{n=0}^{\infty }a_{n}$ and $a_{n}=(-1)^nb_{n}$ or $(-1)^{n-1}b_{n}$ , for the series to converge, $\lim_{n\rightarrow \infty }b_{n}=0$ and $b_{n}$ must be decreasing.

To start, we must take the limit of $b_{n}$ as n approaches infinity:

$\lim_{n\rightarrow \infty }\frac{2n}{n^3}=0$ because $\lim_{n\rightarrow \infty } \frac{n^3}{n^3}(\frac{2n^{-1}}{1})=0$ (the numerator goes to zero).

Next, we must see if $b_{n}$ is decreasing. Simply increasing to n+1 does not clearly show whether the function is decreasing because both the numerator and denominator increase. So, we must find the first derivative and see if it is negative:

$f(n)=\frac{2n}{n^3}$ , $f'(n)=-\frac{4}{n^3}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} (\frac{f(x)}{g(x)})=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$

The derivative is always negative from n=0 to $n= \infty$ , so the sequence $b_{n}$ is decreasing. The series is (absolutely) convergent because it passed both parts of the test.

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Example Question #23 : Alternating Series

Determine whether the series is convergent or divergent:

$\sum_{n=0}^{\infty }\frac{8n^3+7n^2}{6n^3-4n^2}$

Possible Answers:

The series is conditionally convergent

The series is (absolutely) convergent

The series is divergent

The series may be conditionally convergent, (absolutely) convergent, or divergent

Correct answer:

The series is divergent

Explanation:

To determine whether the series is convergent or divergent, we must use the Alternating Series test, which states that for a given series $\sum_{n=0}^{\infty }a_{n}$ , where $a_{n}=(-1)^nb_{n}$ or $(-1)^{n+1}b_{n}$ , if $\lim_{n\rightarrow \infty }b_{n}=0$ and $b_{n}$ is decreasing, then the series is convergent.

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

$\lim_{n\rightarrow\infty }\frac{8n^3+7n^2}{6n^3-4n^2}=\lim_{n\rightarrow \infty }(\frac{n^3}{n^3})(\frac{8+7n^{-1}}{6-4n^{-1}})=\frac{8}{6}\neq0$

Therefore, the test fails and series is divergent.

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Example Question #24 : Alternating Series

Determine whether the following series is convergent or divergent:

$\sum_{n=0}^{\infty }\frac{2n^2-n}{2n^2+5n}$

Possible Answers:

The series is (absolutely) convergent

The series may be (absolutely) convergent, conditionally convergent, or divergent

The series is divergent

The series is conditionally convergent

Correct answer:

The series is divergent

Explanation:

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

$\lim_{n\rightarrow \infty }\frac{2n^2-n}{2n^2+5n}=\lim_{n\rightarrow \infty }\frac{n^2}{n^2}(\frac{2-n^{-1}}{2+5^{-1}})=1\neq0$

The test fails and the series is therefore divergent.

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Example Question #61 : Types Of Series

Determine whether the series is convergent or divergent:

$\sum_{n=0}^{\infty }\frac{n^4}{20n^3+n^2}$

Possible Answers:

The series is (absolutely) convergent

The series is conditionally convergent

The series may be (absolutely) convergent, conditionally convergent, or divergent

The series is divergent

Correct answer:

The series is divergent

Explanation:

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

$\lim_{n\rightarrow \infty }\frac{n^4}{20n^3+n^2}=\frac{n^4}{n^4}(\frac{1}{20n^{-1}+n^{-2}})=\infty \neq0$

The test fails and therefore the series is divergent.

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Example Question #1 : Taylor And Maclaurin Series

For which of the following functions can the Maclaurin series representation be expressed in four or fewer non-zero terms?

Possible Answers:

$f(x)= \frac{x^{5}}{3}+2$

$f(x)=\ln|x+3|$

$f(x)=x+3\sin(x)$

$f(x)= x^{2}+\sqrt{x}+1$

Correct answer:

$f(x)= \frac{x^{5}}{3}+2$

Explanation:

Recall the Maclaurin series formula:

$f(x)=\sum_{n=0}^{\infty}\frac{f^n(0)}{n!}x^n$

$=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+...$

Despite being a 5th degree polynomial recall that the Maclaurin series for any polynomial is just the polynomial itself, so this function's Taylor series is identical to itself with two non-zero terms.

The only function that has four or fewer terms is $f(x)= \frac{x^{5}}{3}+2$ as its Maclaurin series is $2+\frac{1}{3}x^5$ .

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #55 : Types Of Series

Example Question #61 : Types Of Series

Example Question #3021 : Calculus Ii

Example Question #19 : Alternating Series

Example Question #21 : Alternating Series

Example Question #22 : Alternating Series

Example Question #23 : Alternating Series

Example Question #24 : Alternating Series

Example Question #61 : Types Of Series

Example Question #1 : Taylor And Maclaurin Series

All Calculus 2 Resources

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