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Calculus 2 : Series in Calculus

Study concepts, example questions & explanations for Calculus 2

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All Calculus 2 Resources

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

Example Questions

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.

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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.

Report an Error

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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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 #111 : Series In Calculus

Determine the convergence or divergence of the following series:

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

Possible Answers:

The series is conditionally convergent.

The series is (absolutely) convergent.

The series is divergent.

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

Correct answer:

The series is (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, follow the above formula:

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

Now simplify and evaluate the limit:

$\lim_{n\rightarrow \infty }\frac{3(n+5)}{n(n+4)}=\lim_{n\rightarrow \infty }(\frac{3n+15}{n^2+4n})=0<1$

Because the limit is less than one, 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.

Report an Error

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.

Report an Error

Example Question #81 : Convergence And Divergence

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 Neither

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

$\infty$ , and Divergent

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

$\infty$ , 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.

Report an Error

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.

Report an Error

Example Question #111 : 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:

Neither

More tests are needed.

Convergent

Not enough information.

Divergent

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.

Report an Error

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Calculus 2 : Series in Calculus

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #71 : Convergence And Divergence

Example Question #8 : Ratio Test And Comparing Series

Example Question #83 : Ap Calculus Bc

Example Question #10 : Ratio Test And Comparing Series

Example Question #111 : Series In Calculus

Example Question #11 : Ratio Test And Comparing Series

Example Question #112 : Series In Calculus

Example Question #81 : Convergence And Divergence

Example Question #112 : Series In Calculus

Example Question #111 : Series In Calculus

All Calculus 2 Resources

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