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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 #2901 : Calculus Ii

Determine what the following series converges to using the Ratio Test, and whether the series is convergent, divergent or neither.

$\sum_{n=0}^{\infty} \frac{(n+1)!(-5)^n}{n!\2^{n+1}}$

Possible Answers:

$\frac{5}{2}$ , and Convergent

$\infty$ , and Divergent

$\infty$ , and Convergent

$\frac{5}{2}$ , and Divergent

$\frac{5}{2}$ , and Neither

Correct answer:

$\frac{5}{2}$ , and Divergent

Explanation:

To determine if

$\sum_{n=0}^{\infty} \frac{(n+1)!(-5)^n}{n!\2^{n+1}}$

is convergent, divergent or neither, we need to use the ratio test.

The ratio test is as follows.

Suppose we a series $\sum a_n$ . Then we define,

$L=\lim_{n\rightarrow \infty}\left | \frac{a_{n+1}}{a_n}\right |$ .

L<1 the series is absolutely convergent (and hence convergent).

L>1 the series is divergent.

L=1 the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

$a_n=\frac{(n+1)!(-5)^n}{n!\2^{n+1}}$

and

$a_{n+1}=\frac{(n+2)!(-5)^{n+1}}{(n+1)!\2^{n+2}}$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+2)!(-5)^{n+1}}{(n+1)!\2^{n+2}}}{\frac{(n+1)!(-5)^n}{n!\2^{n+1}}}\right |$

We can rearrange the expression to be

$L=\lim_{n\rightarrow \infty}\left | \frac{n!(n+2)!\(-5)^{n+1}\2^{n+1}}{(n+1)!(n+1)!\(-5)^n\2^{n+2}}\right |$

We can simplify the expression to

$L=\frac{5}{2}\lim_{n\rightarrow \infty}\left | \frac{n!(n+2)!}{(n+1)!^2}\right |=\frac{5}{2}\lim_{n\rightarrow \infty}\left | \frac{n+2}{n+1}\right |$

When we evaluate the limit, we get.

$L=\frac{5}{2}$ .

Since L>1 , we have sufficient evidence to conclude that the series is divergent.

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Example Question #112 : Series In Calculus

Determine if the following series is Convergent, Divergent or Neither.

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

Possible Answers:

Divergent

Convergent

Neither

Not enough information.

More tests are needed.

Correct answer:

Divergent

Explanation:

To determine if

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

is convergent, divergent or neither, we need to use the ratio test.

The ratio test is as follows.

Suppose we a series $\sum a_n$ . Then we define,

$L=\lim_{n\rightarrow \infty}\left | \frac{a_{n+1}}{a_n}\right |$ .

L<1 the series is absolutely convergent (and hence convergent).

L>1 the series is divergent.

L=1 the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

$a_n=\frac{(4n+2)(-7)^n}{6^n \ (3n)}$

and

$a_{n+1}=\frac{(4(n+1)+2)(-7)^{n+1}}{6^{n+1} \ (3(n+1))}=\frac{(4n+6)(-7)^{n+1}}{6^{n+1} \ (3n+3)}$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(4n+6)(-7)^{n+1}}{6^{n+1} \ (3n+3)}}{\frac{(4n+2)(-7)^n}{6^n \ (3n)}}\right |$

We can rearrange the expression to be

$L=\lim_{n\rightarrow \infty}\left | \frac{3n(4n+6)(-7)^{n+1}\ 6^n}{(4n+2)(3n+3)6^{n+1}\ (-7)^n}\right |$

We can simplify the expression to

$L=\frac{7}{6} \lim_{n\rightarrow \infty}\left | \frac{3n(4n+6)}{(4n+2)(3n+3)}\right |$

When we evaluate the limit, we get.

$L=\frac{7}{6}$ .

Since L>1 , we have sufficient evidence to conclude that the series is Divergent.

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Example Question #113 : Series In Calculus

Determine if the following series is divergent, convergent or neither by using the ratio test.

$\sum_{n=0}^{\infty} \frac{n}{n+1}$

Possible Answers:

More tests are needed.

Not enough information.

Convergent

Divergent

Neither

Correct answer:

Neither

Explanation:

To determine if

$\sum_{n=0}^{\infty} \frac{n}{n+1}$

is convergent, divergent or neither, we need to use the ratio test.

The ratio test is as follows.

Suppose we a series $\sum a_n$ . Then we define,

$L=\lim_{n\rightarrow \infty}\left | \frac{a_{n+1}}{a_n}\right |$ .

L<1 the series is absolutely convergent (and hence convergent).

L>1 the series is divergent.

L=1 the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

$a_n=\frac{n}{n+1}$

and

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

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{n+1}{n+2}}{\frac{n}{n+1}}\right |$

We can rearrange the expression to be

$L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2}{n(n+2)}\right |$

When we evaluate the limit, we get.

L=1 .

Since L=1 , we have sufficient evidence to conclude that the series neither divergent or convergent.

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Example Question #83 : Convergence And Divergence

Determine if the following series is convergent, divergent or neither.

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

Possible Answers:

Neither

Divergent

More tests are needed.

Not enough information.

Convergent

Correct answer:

Convergent

Explanation:

To determine if

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

is convergent, divergent or neither, we need to use the ratio test.

The ratio test is as follows.

Suppose we a series $\sum a_n$ . Then we define,

$L=\lim_{n\rightarrow \infty}\left | \frac{a_{n+1}}{a_n}\right |$ .

L<1 the series is absolutely convergent (and hence convergent).

L>1 the series is divergent.

L=1 the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

$a_n=\frac{n!\ 2^n}{(n+1)!\ 3^n}$

and

$a_{n+1}=\frac{(n+1)!\ 2^{n+1}}{(n+2)!\ 3^{n+1}}$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)!\ 2^{n+1}}{(n+2)!\ 3^{n+1}}}{\frac{n!\ 2^n}{(n+1)!\ 3^n}}\right |$

We can rearrange the expression to be

$L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)!(n+1)!\2^{n+1}\3^n}{n!(n+2)!\2^n\3^{n+1}}\right |$

We can simplify the expression to

$L=\frac{2}{3}\lim_{n\rightarrow \infty}\left | \frac{(n+1)!^2}{n!(n+2)!}\right |=\frac{2}{3}\lim_{n\rightarrow \infty}\left | \frac{n+1}{n+2}\right |$

When we evaluate the limit, we get.

$L=\frac{2}{3}$ .

Since L<1 , we have sufficient evidence to conclude that the series is convergent.

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Example Question #2902 : Calculus Ii

Determine if the following series is Convergent, Divergent or Neither.

$\sum_{n=0}^{\infty} \frac{(2n+3)4^{n+2}}{(n+2)5^{n+1}}$

Possible Answers:

Not enough information.

Neither

Convergent

More tests are needed.

Divergent

Correct answer:

Convergent

Explanation:

To determine if

$\sum_{n=0}^{\infty} \frac{(2n+3)4^{n+2}}{(n+2)5^{n+1}}$

is convergent, divergent or neither, we need to use the ratio test.

The ratio test is as follows.

Suppose we a series $\sum a_n$ . Then we define,

$L=\lim_{n\rightarrow \infty}\left | \frac{a_{n+1}}{a_n}\right |$ .

L<1 the series is absolutely convergent (and hence convergent).

L>1 the series is divergent.

L=1 the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

$a_n=\frac{(2n+3)4^{n+2}}{(n+2)5^{n+1}}$

and

$a_{n+1}=\frac{(2(n+1)+3)4^{n+3}}{(n+3)5^{n+2}}=\frac{(2n+5)4^{n+3}}{(n+3)5^{n+2}}$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(2n+5)4^{n+3}}{(n+3)5^{n+2}}}{\frac{(2n+3)4^{n+2}}{(n+2)5^{n+1}}}\right |$

We can rearrange the expression to be

$L=\lim_{n\rightarrow \infty}\left | \frac{(2n+5)(n+2)4^{n+3}\ 5^{n+1}}{(n+3)(2n+3)5^{n+2}\ 4^{n+2}}\right |$

We can simplify the expression to

$L=\frac{4}{5}\lim_{n\rightarrow \infty}\left | \frac{(2n+5)(n+2)}{(n+3)(2n+3)}\right |$

When we evaluate the limit, we get.

$L=\frac{4}{5}$ .

Since L<1 , we have sufficient evidence to conclude that the series is convergent.

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Example Question #85 : Convergence And Divergence

Determine whether the following series is Convergent, Divergent or Neither.

$\sum_{n=0}^{\infty} \frac{n2^n}{n!}$

Possible Answers:

Divergent

Neither

Not enough information.

More tests are needed.

Convergent

Correct answer:

Convergent

Explanation:

To determine if

$\sum_{n=0}^{\infty} \frac{n2^n}{n!}$

is convergent, divergent or neither, we need to use the ratio test.

The ratio test is as follows.

Suppose we a series $\sum a_n$ . Then we define,

$L=\lim_{n\rightarrow \infty}\left | \frac{a_{n+1}}{a_n}\right |$ .

L<1 the series is absolutely convergent (and hence convergent).

L>1 the series is divergent.

L=1 the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

$a_n=\frac{n2^n}{n!}$

and

$a_{n+1}=\frac{(n+1)2^{n+1}}{(n+1)!}$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)2^{n+1}}{(n+1)!}}{\frac{n2^n}{n!}}\right |$

We can rearrange the expression to be

$L=\lim_{n\rightarrow \infty}\left | \frac{n!(n+1)\ 2^{n+1}}{n(n+1)!\ 2^{n}}\right |$

We can simplify the expression to

$L=2\lim_{n\rightarrow \infty}\left | \frac{1}{n}\right |$

When we evaluate the limit, we get.

L=0 .

Since L<1 , we have sufficient evidence to conclude that the series is convergent.

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Example Question #121 : Series In Calculus

Determine whether the following series converges or diverges:

$\sum_{n=0}^{\infty }\frac{2^n}{(n+2)!}$

Possible Answers:

The series diverges.

The series (absolutely) converges.

The series conditionally converges.

The series may converge, diverge, or be conditionally convergent.

Correct answer:

The series (absolutely) converges.

Explanation:

To determine whether the given series is convergent or divergent, we must use the Ratio Test, which states that for

$L=\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_{n}}\right |$

if L is less than 1, then the series is absolutely convergent, and therefore convergent;

if L is greater than 1, then the series is divergent;

if L is equal to 1, the series may be convergent, divergent, or conditionally convergent.

Now, use the above formula and evaluate the limit:

$\lim_{n\rightarrow \infty }\left | \frac{2^{n+1}}{(n+3)(n+2)!}\cdot \frac{(n+2)!}{2^n}\right |$

which simplified becomes

$\lim_{n\rightarrow \infty }\left | \frac{2}{n+3}\right |=0<1$

Because L is less than 1, the series is (absolutely) convergent.

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Example Question #2901 : Calculus Ii

Consider the series

$\sum_{n=0}^{\infty} \frac{1}{n^2+10}$ .

Using the ratio test, what can we conclude regarding its convergence?

Possible Answers:

We can conclude nothing using the ratio test. Another test will have to be used to test its convergence.

Ratio test implies that the series is conditionally convergent.

Ratio test implies that the series is absolutely convergent.

Ratio test implies that the series is divergent.

Correct answer:

We can conclude nothing using the ratio test. Another test will have to be used to test its convergence.

Explanation:

Let's use the ratio test to see what we can conclude:

$\lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n} \right |=\lim_{n\to \infty} \left|\frac{1}{(n+1)^2+10}\frac{n^2+10}{1} \right |=1$

Since the ratio test limit results in one, we cannot conclude anything about the series' convergence/divergence by definition of the ratio test.

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Example Question #123 : Series In Calculus

Find the radius of convergence of the following power series:

$\sum_{n=0}^{\infty} \frac{x^{n}}{(2n)!}$

Possible Answers:

$\infty$

Correct answer:

$\infty$

Explanation:

Let's use the ratio test to find the radius of convergence of

$\sum_{n=0}^{\infty} \frac{x^{n}}{(2n)!}$

$\lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n} \right |=\lim_{n\to \infty} \left|\frac{x^{n+1}}{(2n+2)!}\frac{(2n)!}{x^n} \right |=\lim_{n\to \infty} \left|\frac{x}{(2n+1)(2n+2)} \right |$

independent of . This means the series converges for all , so the radius of convergence is $\infty$ .

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Example Question #124 : Series In Calculus

Find the radius of convergence of the power series.

$\sum_{n=0}^{\infty} \frac{x^{7n}(2n)!}{n!n!}$

Possible Answers:

$R=\frac{1}{4^{\frac{1}{7}}}$

R=0

$R=\frac{1}{4^7}$

$R=4^{\frac{1}{7}}$

Correct answer:

$R=\frac{1}{4^{\frac{1}{7}}}$

Explanation:

We can find the radius of convergence of

$\sum_{n=0}^{\infty} \frac{x^{7n}(2n)!}{n!n!}$ using the ratio test:

$\lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n} \right |=\lim_{n\to \infty} \left|\frac{x^{7n+7}(2n+2)!}{(n+1)!(n+1)!}\frac{n!n!}{x^{7n}(2n)!} \right |=\lim_{n\to \infty} \left|\frac{x^7(2n+1)(2n+2)}{(n+1)(n+1)} \right |$

$=\left|4x^7 \right |<1$

So then we have

$|x|<\frac{1}{4^{\frac{1}{7}}}=R$

Which means

$R=\frac{1}{4^{\frac{1}{7}}}$

is the radius of convergence.

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2901 : Calculus Ii

Example Question #112 : Series In Calculus

Example Question #113 : Series In Calculus

Example Question #83 : Convergence And Divergence

Example Question #2902 : Calculus Ii

Example Question #85 : Convergence And Divergence

Example Question #121 : Series In Calculus

Example Question #2901 : Calculus Ii

Example Question #123 : Series In Calculus

Example Question #124 : Series In Calculus

All Calculus 2 Resources

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