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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 #1 : P Series

True or False, a -series cannot be tested conclusively using the ratio test.

Possible Answers:

False

True

Correct answer:

True

Explanation:

We cannot test for convergence of a -series using the ratio test. Observe,

For the series $\sum_{n=1}^\infty \frac{1}{n^p}$ ,

$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} | \frac{n^p}{(n+1)^p}| = \lim_{n \to \infty} (\frac{n}{n+1})^p = 1^p =1$ .

Since this limit is regardless of the value for , the ratio test is inconclusive.

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Example Question #2881 : Calculus Ii

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

$\sum_{n=1}^{\infty} \frac{(-9)^n}{5^{3n+2}(n+1)}$

Possible Answers:

Divergent

Convergent

Neither

Both

Inconclusive

Correct answer:

Convergent

Explanation:

In order to figure out if

$\sum_{n=1}^{\infty} \frac{(-9)^n}{5^{3n+2}(n+1)}$

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

Remember that the ratio test is as follows.

Suppose we have a series $\sum a_n$ . We define,

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

Then if

L<1 , the series is absolutely convergent.

L>1 , the series is divergent.

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

Now lets apply the ratio test to our problem.

Let

$a_n=\frac{(-9)^n}{5^{3n+2}(n+1)}$

and

$a_{n+1}=\frac{(-9)^{n+1}}{5^{3(n+1)+2}((n+1)+1)}$

$=\frac{(-9)^{n+1}}{5^{3n+5}(n+2)}$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(-9)^{n+1}}{5^{3n+5}(n+2)}}{\frac{(-9)^n}{5^{3n+2}(n+1)}} \right |$

$=\lim_{n\rightarrow \infty}\left | \frac{(-9)^{n+1}5^{3n+2}(n+1)}{(-9)^n5^{3n+5}(n+2)} \right |$

Now lets simplify this expression to

$=\lim_{n\rightarrow \infty}\left | \frac{(-9)^{1}(n+1)}{5^{3}(n+2)} \right |$

$=\frac{9}{125}\lim_{n\rightarrow \infty}\left | \frac{n+1}{n+2} \right |$

$=\frac{9}{125}*1=\frac{9}{125}$ .

Since

$\frac{9}{125}<1$ .

We have sufficient evidence to conclude that the series is convergent.

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Example Question #81 : Polynomial Approximations And Series

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

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

Possible Answers:

Neither

Inconclusive

Convergent

Divergent

Both

Correct answer:

Divergent

Explanation:

In order to figure if

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

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

Remember that the ratio test is as follows.

Suppose we have a series $\sum a_n$ . We define,

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

Then if

L<1 , the series is absolutely convergent.

L>1 , the series is divergent.

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

Now lets apply the ratio test to our problem.

Let

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

and

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

Now

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

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

Now lets simplify this expression to

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

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

$=\frac{1}{2}*\infty=\infty$ .

Since $\infty>1$ ,

we have sufficient evidence to conclude that the series is divergent.

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Example Question #2883 : Calculus Ii

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

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

Possible Answers:

Inconclusive

Convergent

Neither

Both

Divergent

Correct answer:

Divergent

Explanation:

In order to figure if

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

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

Remember that the ratio test is as follows.

Suppose we have a series $\sum a_n$ . We define,

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

Then if

L<1 , the series is absolutely convergent.

L>1 , the series is divergent.

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

Now lets apply the ratio test to our problem.

Let

$a_n=\frac{11^n}{(n+1)3^{n}}$

and

$a_{n+1}=\frac{11^{n+1}}{(n+2)3^{n+1}}$ .

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{11^{n+1}}{(n+2)3^{n+1}}}{\frac{11^n}{(n+1)3^{n}}} \right |$

$=\lim_{n\rightarrow \infty}\left | \frac{(n+1)11^{n+1}3^n}{11^{n}(n+2)3^{n+1}} \right |$ .

Now lets simplify this expression to

$=\lim_{n\rightarrow \infty}\left | \frac{11(n+1)}{3(n+2)} \right |$

$=\frac{11}{3}\lim_{n\rightarrow \infty}\left |\frac{(n+1)}{(n+2)} \right |$

$=\frac{11}{3}*1=\frac{11}{3}$ .

Since $\frac{11}{3}>1$ ,

we have sufficient evidence to conclude that the series is divergent.

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Example Question #4 : Ratio Test And Comparing Series

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

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

Possible Answers:

Divergent

Convergent

Neither

More tests are needed.

Inconclusive

Correct answer:

Divergent

Explanation:

To determine if

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

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 therefore 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

$a_n=\frac{n!\ 4^n}{n+1}$

and

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

Now

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

We can rearrange the expression to be

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

Now lets simplify this.

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

When we evaluate the limit, we get.

$L=\infty$ .

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

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Example Question #5 : Ratio Test And Comparing Series

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

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

Possible Answers:

More tests are needed.

Divergent

Inconclusive

Neither

Convegent

Correct answer:

Convegent

Explanation:

To determine if

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

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 thus 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

$a_n=\frac{n \ 2^n}{5^n}$

and

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

Now

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

We can rearrange the expression to be

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

Now lets simplify this.

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

When we evaluate the limit, we get.

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

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

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Example Question #6 : Ratio Test And Comparing Series

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

$\sum_{n=1}^{\infty} \frac{n(-100)^n}{(n+1)5^{n+2}}$

Possible Answers:

Neither

Inconclusive

Convergent

Divergent

More tests needed.

Correct answer:

Divergent

Explanation:

To determine if

$\sum_{n=1}^{\infty} \frac{n(-100)^n}{(n+1)5^{n+2}}$

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 (therefore 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

$a_n=\frac{n(-100)^n}{(n+1)5^{n+2}}$

and

$a_{n+1}=\frac{(n+1)(-100)^{n+1}}{(n+2)5^{n+3}}$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)(-100)^{n+1}}{(n+2)5^{n+3}}}{\frac{n(-100)^n}{(n+1)5^{n+2}}}\right |$

We can rearrange the expression to be

$L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2(-100)^{n+1}{5^{n+2}}}{n(n+2)\ (-100)^n \ 5^{n+3}}\right |$

Now lets simplify this.

$L=20\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2}{n(n+2)}\right |$

When we evaluate the limit, we get.

L=20 .

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

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Example Question #71 : Convergence And Divergence

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

$\sum_{n=0}^{\infty} 4^{2n+1}n!$

Possible Answers:

Neither

Divergent

Convergent

Inconclusive

More tests are needed.

Correct answer:

Divergent

Explanation:

To determine if

$\sum_{n=0}^{\infty} 4^{2n+1}n!$

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 thus 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

$a_n= 4^{2n+1}n!$

and

$a_{n+1}= 4^{2(n+1)+1}(n+1)!=4^{2n+3}(n+1)!$

Now

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

We can simplify the expression to be

$L=\lim_{n\rightarrow \infty}\left | 16 (n+1)}\right |$

When we evaluate the limit, we get.

$L=\infty$ .

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

Report an Error

Example Question #8 : Ratio Test And Comparing Series

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

$\sum_{n=0}^{\infty} \frac{n4^{2n}}{(n+1)3^{2n}}$

Possible Answers:

Divergent

Inconclusive.

Convergent

Neither

More tests are needed.

Correct answer:

Divergent

Explanation:

To determine whether this series is convergent, divergent or neither

$\sum_{n=0}^{\infty} \frac{n4^{2n}}{(n+1)3^{2n}}$

we need to remember 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 therefore 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

$a_n= \frac{n4^{2n}}{(n+1)3^{2n}}$

and

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

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)4^{2n+2}}{(n+2)3^{2n+2}}}{\frac{n4^{2n}}{(n+1)3^{2n}}}\right |$

We can rearrange the expression to be

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

Now lets simplify this to.

$L=\frac{16}{9}\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2}{n(n+2)}\right |$

When we evaluate the limit, we get.

$L=\frac{16}{9}$ .

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

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Example Question #83 : Ap Calculus Bc

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

$\sum_{n=0}^{\infty} \frac{n2^{4n}}{(n+1)5^{6n}}$

Possible Answers:

$L=\frac{16}{15625}$ , and neither.

L=1 , and neither.

$L=\frac{16}{15625}$ , and divergent.

L=1 , and convergent.

$L=\frac{16}{15625}$ , and convergent.

Correct answer:

$L=\frac{16}{15625}$ , and convergent.

Explanation:

To determine whether this series is convergent, divergent or neither

$\sum_{n=0}^{\infty} \frac{n2^{4n}}{(n+1)5^{6n}}$

we need to remember 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 (thus 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

$a_n= \frac{n2^{4n}}{(n+1)5^{6n}}$

and

$a_{n+1}= \frac{(n+1)2^{4(n+1)}}{(n+2)5^{6(n+1)}}= \frac{(n+1)2^{4n+4}}{(n+2)5^{6n+6}}$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)2^{4n+4}}{(n+2)5^{6n+6}}}{\frac{n2^{4n}}{(n+1)5^{6n}}}\right |$

We can rearrange the expression to be

$L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2\ 2^{4n+4} \ 5^{6n}}{n(n+2)\ 2^{4n}\ 5^{6n+6}}\right |$

Now lets simplify this to.

$L=\frac{16}{15625}\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2}{n(n+2)}\right |$

When we evaluate the limit, we get.

$L=\frac{16}{15625}$ .

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

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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 #1 : P Series

Example Question #2881 : Calculus Ii

Example Question #81 : Polynomial Approximations And Series

Example Question #2883 : Calculus Ii

Example Question #4 : Ratio Test And Comparing Series

Example Question #5 : Ratio Test And Comparing Series

Example Question #6 : Ratio Test And Comparing Series

Example Question #71 : Convergence And Divergence

Example Question #8 : Ratio Test And Comparing Series

Example Question #83 : Ap Calculus Bc

All AP Calculus BC Resources

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