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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 #91 : Series In Calculus

Use the ratio test to find out if the following series is convergent:

$\sum_{n = 0}^{\infty} \frac{6n!}{6^n\left(6n-1\right)!} \newline note: \hspace f(n) = \frac{6n!}{6^n\left(6n-1\right)!}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{1}{6}$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 6$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

Explanation:

Determine the convergence of the series based on the limits.

$\newline Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} < 1 \newline Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} > 1 \newline Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take $\lim_{n\rightarrow \infty}$ .

$\frac{\frac{\left(n+1\right)!}{6^{\left(n+1\right)}\left(6\left(n+1\right)-1\right)!}}{\frac{n!}{6^n\left(6n-1\right)!}} = \frac{6^n\left(n+1\right)!\left(6n-1\right)!}{6^{n+1}n!\left(6\left(n+1\right)-1\right)!}$ $= \frac{\left(6n-1\right)!\left(n+1\right)}{6\left(6\left(n+1\right)-1\right)!} = \frac{\left(6n-1\right)!\left(n+1\right)}{6\left(6n+5\right)!}$

$\lim_{n\rightarrow \infty} \frac{\left(6n-1\right)!\left(n+1\right)}{6\left(6n+5\right)!} = 0$

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Example Question #92 : Series In Calculus

Use the ratio test to find out if the following series is convergent:

$\sum_{n = 0}^{\infty} \frac{n!}{3^n\left(2n-1\right)!} \newline note: \hspace f(n) = \frac{n!}{3^n\left(2n-1\right)!}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{1}{3}$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 5$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

Explanation:

Determine the convergence of the series based on the limits.

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take $\lim_{n\rightarrow \infty}$ .

$\frac{\frac{\left(n+1\right)!}{3^{\left(n+1\right)}\left(2\left(n+1\right)-1\right)!}}{\frac{n!}{3^n\left(2n-1\right)!}} = \frac{3^n\left(n+1\right)!\left(2n-1\right)!}{3^{n+1}n!\left(2\left(n+1\right)-1\right)!}$

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

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

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Example Question #93 : Series In Calculus

Use the ratio test to find out if the following series is convergent:

$\sum_{n = 0}^{\infty} \frac{n!}{2^n\left(8n-10\right)!} \newline note: \hspace f(n) = \frac{n!}{2^n\left(8n-10\right)!}$

Possible Answers:

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{1}{2}$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 8$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

Explanation:

Determine the convergence of the series based on the limits.

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take $\lim_{n\rightarrow \infty}$ .

$\frac{\frac{\left(n+1\right)!}{2^{\left(n+1\right)}\left(8\left(n+1\right)-10\right)!}}{\frac{n!}{2^n\left(8n-10\right)!}} = \frac{2^n\left(n+1\right)!\left(8n-10\right)!}{2^{n+1}n!\left(8\left(n+1\right)-10\right)!}$

$= \frac{\left(8n-10\right)!\left(n+1\right)}{2\left(8n-2\right)!}$

$\lim_{n\rightarrow \infty} \frac{\left(8n-10\right)!\left(n+1\right)}{2\left(8n-2\right)!} = 0$

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Example Question #101 : Series In Calculus

Use the ratio test to find out if the following series is convergent:

$\sum_{n = 0}^{\infty} \frac{n!}{3^n\left(n-1\right)!} \newline note: \hspace f(n) = \frac{n!}{3^n\left(n-1\right)!}$

Possible Answers:

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 3$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{1}{3}$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{1}{3}$

Explanation:

Determine the convergence of the series based on the limits.

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take $\lim_{n\rightarrow \infty}$ .

$\frac{\frac{\left(n+1\right)!}{3^{\left(n+1\right)}\left(\left(n+1\right)-1\right)!}}{\frac{n!}{3^n\left(n-1\right)!}} = \frac{3^n\left(n+1\right)!\left(n-1\right)!}{3^{n+1}n!n!}$

$= \frac{3^n\left(n+1\right)!\left(n-1\right)!}{3^{n+1}n!^2} = \frac{\left(n+1\right)!\left(n-1\right)!}{3n!^2}$

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

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Example Question #102 : Series In Calculus

Use the ratio test to find out if the following series is convergent:

$\sum_{n = 0}^{\infty} \frac{3^n\cdot \:4n!}{\left(n-1\right)!} \newline note: \hspace f(n) = \frac{3^n\cdot \:4n!}{\left(n-1\right)!}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0$

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = \frac{1}{4}$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} \rightarrow \infty$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 3$

$Inconclusive: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1$

Correct answer:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 3$

Explanation:

Determine the convergence of the series based on the limits.

Solution:

1. Ignore constants and simplify the equation (canceling out what you can).

2. Once the equation is simplified, take $\lim_{n\rightarrow \infty}$ .

$\frac{\frac{3^{n+1}\left(n+1\right)!}{\left(\left(n+1\right)-1\right)!}}{\frac{3^nn!}{\left(n-1\right)!}} = \frac{3^{n+1}\left(n+1\right)!\left(n-1\right)!}{3^nn!^2}$ $= \frac{3\left(n+1\right)!\left(n-1\right)!}{n!^2}$

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

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Example Question #1 : Ratio Test And Comparing Series

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

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

Possible Answers:

Convergent

Both

Neither

Inconclusive

Divergent

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 #2 : Ratio Test And Comparing Series

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

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

Possible Answers:

Neither

Inconclusive

Both

Convergent

Divergent

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 #3 : Ratio Test And Comparing Series

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

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

Possible Answers:

Divergent

Inconclusive

Both

Neither

Convergent

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 #104 : Series In Calculus

Determine if the series converges or diverges:

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

Possible Answers:

Neither converges nor diverges.

There is not enough information to determine convergency.

Conditionally converges.

Diverges

Converges

Correct answer:

Converges

Explanation:

The ratio test states that if you take the n+1 term of the series and divide it by the n term, and then take the limit as n approaches infinity and if you take the absolute value of your answer and if it less than 1, it converges.

If it is greater than 1, it diverges.

If it is 1, the test is inconclusive.

The n+1 term is $\frac{(2n+2)!}{(3n+3)!}$ .

Note that you are substituting n+1 for n and so you will distribute the term by 2 and 3 respectively. Dividing the n+1 term by the n term gives you the following: $\frac{(2n+2)!}{(3n+3)!} \cdot \frac{(3n)!}{(2n)!}$ , which when multiplied out gives us the following:

$\frac{(2n+1)(2n+2)}{(3n+1)(3n+2)(3n+3)}$ .

To use the ratio test, we must take the limit of this term as n approaches infinity. From inspection, we can see that the denominator is increasing much faster than the numerator (there are more n terms) and so the limit as n appraoches infinity is 0. Since the absolute value of 0 is less than 1, the series converges.

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Example Question #67 : Convergence And Divergence

Determine what the limit is using the Ratio Test.

$\sum_{n=0}^{\infty} \frac{n5^{n+1}}{(n+1)4^n}$

Possible Answers:

$\frac{5}{4}$

$\frac{4}{5}$

$\infty$

Correct answer:

$\frac{5}{4}$

Explanation:

To determine what this series converges to

$\sum_{n=0}^{\infty} \frac{n5^{n+1}}{(n+1)4^n}$

we need to remember the ratio test.

The ratio test is as follows.

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

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

Now lets apply this to our situtation.

Let

$a_n=\frac{n5^{n+1}}{(n+1)4^n}$

and

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

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)5^{n+2}}{(n+2)4^{n+1}}}{\frac{n5^{n+1}}{(n+1)4^n}}\right |$

We can rearrange the expression to be

$L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)^25^{n+2}4^n}{(n+2)5^{n+1}4^{n+1}}\right |$

Now lets simplify this.

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

When we evaluate the limit, we get.

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

Thus the limit of this series is $\frac{5}{4}$ using the ratio test.

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #91 : Series In Calculus

Example Question #92 : Series In Calculus

Example Question #93 : Series In Calculus

Example Question #101 : Series In Calculus

Example Question #102 : Series In Calculus

Example Question #1 : Ratio Test And Comparing Series

Example Question #2 : Ratio Test And Comparing Series

Example Question #3 : Ratio Test And Comparing Series

Example Question #104 : Series In Calculus

Example Question #67 : Convergence And Divergence

All Calculus 2 Resources

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