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AP Calculus BC : AP Calculus BC

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 #10 : Ratio Test And Comparing Series

Determine the convergence or divergence of the following series:

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

Possible Answers:

The series is divergent.

The series is conditionally convergent.

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

The series (absolutely) convergent.

Correct answer:

The series (absolutely) convergent.

Explanation:

To determine the convergence or divergence of this series, we use the Ratio Test:

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

If L<1 , then the series is absolutely convergent (convergent)

If L>1 , then the series is divergent

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

So, we evaluate the limit according to the formula above:

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

which simplified becomes

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

Further simplification results in

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

Therefore, the series is absolutely convergent.

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Example Question #11 : Ratio Test And Comparing Series

Using the Ratio Test, determine what the following series converges to, and whether the series is Divergent, Convergent or Neither.

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

Possible Answers:

L=2 , and Convergent

$L=\infty$ , and Neither

L=2 , and Neither

$L=\infty$ , and Divergent

L=2 , and Divergent

Correct answer:

L=2 , and Divergent

Explanation:

To determine if

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

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

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

and

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

Now

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

We can rearrange the expression to be

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

We can simplify the expression to

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

When we evaluate the limit, we get.

L=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 what the following series converges to, and whether the series is Convergent, Divergent or Neither.

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

Possible Answers:

$L=\frac{2}{3}$ , and Divergent

$L=\frac{2}{3}$ , and Neither

$L=\frac{2}{3}$ , and Convergent

L=1 , and Neither

$L=\infty$ , and Convergent

Correct answer:

$L=\frac{2}{3}$ , and Convergent

Explanation:

To determine if

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

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

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

and

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

Now

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

We can rearrange the expression to be

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

We can simplify the expression to

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

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 #91 : Ap Calculus Bc

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 Divergent

$\infty$ , and Divergent

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

$\infty$ , and Convergent

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

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 #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 #2 : Comparing Series

We know that :
$chx=\frac{e^x+e^{-x}}{2}$ and $shx=\frac{e^x-e^{-x}}{2}$

We consider the series having the general term:

$v_{n}=\frac{chn}{ch2n}$

Determine the nature of the series:

$\sum_{n=1}^{\infty}v_{n}$

Possible Answers:

$e^{-4}$

$\frac{1}{2}$

The series is divergent.

It will stop converging after a certain number.

The series is convergent.

Correct answer:

The series is convergent.

Explanation:

We know that:

$chn=\frac{e^n+e^{-n}}{2}$ and therefore we deduce :

$ch2n=\frac{e^{2n }+e^{-2n}}{2}$

We will use the Comparison Test with this problem. To do this we will look at the function in general form $v_{n}\equiv \frac{e^n} {e^{2n}}$ .

We can do this since,

$e^{-n}=\frac{1}{e^n}$ and $e^{-2n}=\frac{1}{e^{2n}}$ approach zero as n approaches infinity. The limit of our function becomes,

$\lim_{n\rightarrow \infty}\frac{e^n+e^{-n}}{e^{2n}+e^{-2n}}=\frac{e^n}{e^{2n}}$

This last part gives us $v_{n}\equiv {e^{-n}}$ .

Now we know that $\frac{1}{e}< 1$ and noting that $e^{-n}}$ is a geometric series that is convergent.

We deduce by the Comparison Test that the series

having general term $v_{n}$ is convergent.

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Example Question #3 : Comparing Series

We consider the following series:

$\sum_{n=0}^{\infty} \frac{n^3+1}{n^4}$$

Determine the nature of the convergence of the series.

Possible Answers:

The series is divergent.

$\frac{3}4{}$

$\frac{1}{2}$

Correct answer:

The series is divergent.

Explanation:

We will use the comparison test to prove this result. We must note the following:

$\frac{n^3+1}{n^4}$ is positive.

We have all natural numbers n:

$n^4\ge n^3$ , this implies that

$n^4\ge n^3 \ge n$ .

Inverting we get :

$\frac{1}{n} <\frac{n^3+1}{n^4}$

Summing from 1 to $\infty$ , we have

$\sum_{n=1}^{\infty}\frac{1}{n} < \sum_{n=1}^{\infty}\frac{n^3+1}{n^4}$

We know that the $\sum_{n=1}^{\infty}\frac{1}{n}$ is divergent. Therefore by the comparison test:

$\sum_{n=1}^{\infty}\frac{n^3}{n^4+1}$

is divergent

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Example Question #121 : Convergence And Divergence

Using the Limit Test, determine the nature of the series:

$\frac{n+1}{n^9-7}$

Possible Answers:

$\frac{1}{5}$

$\frac{2}{7}$

The series is divergent.

$\frac{1}{99}$

The series is convergent.

Correct answer:

The series is convergent.

Explanation:

We will use the Limit Comparison Test to study the nature of the series.

We note first that $n\ge 2$ , the series is positive.

We will compare the general term to $\frac{1}{n^8}$ .

We note that by letting $v_{n}=\frac{n+11}{n^9-7}$ and $u_{n}=\frac{1}{n^8}$ , we have:

$\lim_{n\rightarrow \infty}\frac{u_{n}}{v_n}}=1$ .

Therefore the two series have the same nature, (they either converge or diverge at the same time).

We will use the Integral Test to deduce that the series having the general term:

$u_{n}=\frac{1}{n^8}$ is convergent.

Note that we know that $\int_{1}^{\infty}\frac{1}{x^p}dx$ is convergent if p>1 and in our case p=8 .

This shows that the series having general term $u_{n}$ is convergent.

By the Limit Test, the series having general term v_n is convergent.

This shows that our series is convergent.

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Example Question #123 : Convergence And Divergence

We consider the following series:

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

Determine the nature of the convergence of the series.

Possible Answers:

$\frac{3}4{}$

$\frac{1}{2}$

The series is divergent.

Correct answer:

The series is divergent.

Explanation:

We will use the Comparison Test to prove this result. We must note the following:

$\frac{n}{n^2+1}$ is positive.

We have all natural numbers n:

$n^2\ge n$ , this implies that

$n^2+1\ge n+1 > n$ .

Inverting we get :

$\frac{1}{n} <\frac{n}{n^2+1}$

Summing from 1 to $\infty$ , we have

$\sum_{n=1}^{\infty}\frac{1}{n} < \sum_{n=1}^{\infty}\frac{n}{n^2+1}$

We know that the $\sum_{n=1}^{\infty}\frac{1}{n}$ is divergent. Therefore by the Comparison Test:

$\sum_{n=1}^{\infty}\frac{n}{n^2+1}$ is divergent.

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Example Question #162 : Series In Calculus

Is the series

$\small \sum_{n=1}^\infty \frac{n^3}{n^7+1000}$

convergent or divergent, and why?

Possible Answers:

Divergent, by the comparison test.

Convergent, by the ratio test.

Convergent, by the comparison test.

Divergent, by the test for divergence.

Divergent, by the ratio test.

Correct answer:

Convergent, by the comparison test.

Explanation:

We will use the comparison test to prove that

$\small \sum_{n=1}^\infty \frac{n^3}{n^7+1000}$

converges (Note: we cannot use the ratio test, because then the ratio will be $\small 1$ , which means the test is inconclusive).

We will compare $\small \frac{n^3}{n^7+1000}$ to $\small \frac{1}{n^4}$ because they "behave" somewhat similarly. Both series are nonzero for all $\small n\geq 1$ , so one of the conditions is satisfied.

The series

$\small \small \sum_{n=1}^\infty \frac{1}{n^4}$

converges, so we must show that

$\small \small \frac{1}{n^4}\geq \frac{n^3}{n^4+1000}$

for $\small n\geq 1$ .

This is easy to show because

$\small \frac{1}{n^4}=\frac{n^3}{n^7}\geq \frac{n^3}{n^7+1000}$

since the denominator $\small n^7+1000$ is greater than or equal to $\small n^7$ for all $\small n\geq 1$ .

Thus, since

$\small \small \frac{1}{n^4}\geq \frac{n^3}{n^4+1000}$

and because

$\small \small \sum_{n=1}^\infty \frac{1}{n^4}$

converges, it follows that

$\small \sum_{n=1}^\infty \frac{n^3}{n^7+1000}$

converges, by comparison test.

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AP Calculus BC : AP Calculus BC

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #10 : Ratio Test And Comparing Series

Example Question #11 : Ratio Test And Comparing Series

Example Question #112 : Series In Calculus

Example Question #91 : Ap Calculus Bc

Example Question #2902 : Calculus Ii

Example Question #2 : Comparing Series

Example Question #3 : Comparing Series

Example Question #121 : Convergence And Divergence

Example Question #123 : Convergence And Divergence

Example Question #162 : Series In Calculus

All AP Calculus BC Resources

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