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Calculus 2 : Ratio Test

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Example Questions

Example Question #41 : Series In Calculus

Let the series $\sum_{n=1}^{\infty}a_{n}$ with $a_{n}=\frac{(-1)^{n+1}}{n}$ determine whether the series is convergent or divergent using the ratio test.

Possible Answers:

We can't conclude when using the ratio test.

$\frac{-1}{ln2}$

We cant use the ratio test for this type of series.

The series is divergent.

The series is convergent.

Correct answer:

We can't conclude when using the ratio test.

Explanation:

Note that the series is alternating. To be able to use the ratio test, we will have to compute the ratio:

$\left|\frac{a_{n+1}}{a_n}\ \right|=\frac{(-1)^{n+2}}{n+1}\cdot \frac{n}{(-1)^{n+1}}=\frac{n}{n+1}$

Now we need to see that :

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

Since $\lim_{n\rightarrow \infty }\left|\frac{a_{n+1}}{a_n}\ \right|=1$ we can't conclude by using the ratio test. We will have to call upon another test to show that the series is convergent.

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Example Question #11 : Ratio Test

Determine if the statement is true or false.

Assume that the series has positive terms. Furthermore suppose that

$\lim _{n\rightarrow \infty}\frac{a_{n+1}}{a_n}=1$ , then all the series of the form $\sum_{n=1}^{\infty}a_n$ are divergent.

Possible Answers:

The series is convergent if it is positive.

There is only one series that satisfies that.

The series is divergent.

We can't conclude in general.

The statement above is false.

Correct answer:

The statement above is false.

Explanation:

To show that the statement above is false, we will consider the following example.

consider the series $\sum_{n=1}^{\infty}a_{n}$ where $a_{n}$ is given by:

$a_{n}=\frac{1}{n(n+1)}$ clearly the series has positive terms. Furthermore, we have

$\lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n}=\frac{1}{(n+1)(n+2)}{\cdot \frac{n(n+1)}{1}}=\frac{n}{n+2}=1$ , meaning the series can be either convergent or divergent.

However, the series : $\sum_{n=1}^{\infty}a_{n}$ is convergent (use for example the integral test to see that it is convergent).

Therefore, the original statement is false.

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Example Question #11 : Ratio Test

We consider the series,

$\sum_{n=1}^{\infty}\frac{a^n}{n^b},0<a<\frac{1}{999} ,b>0$ .

Using the ratio test, what can we conclude about the nature of convergence of this series?

Possible Answers:

We will need to know the values of to decide.

The series is convergent.

We can't use the ratio test here.

The series is divergent.

The series converges to $\frac{a}{b}$ .

Correct answer:

The series is convergent.

Explanation:

Note that the series is positive.

As it is required we will use the ratio test to check for the nature of the series.

We have $\frac{a_{n+1}}{a_n}=\frac{a^{n+1}}{(n+1)^b}\frac{n^b}{a^n}=a\left(\frac{n}{n+1}\right)^b$ .

Therefore, $\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\lim_{n\rightarrow\infty} a(\frac{n}{n+1})b$ $\lim_{n\rightarrow\infty}\frac{n^b}{(n+1)^b}=a(\lim_{n\rightarrow}\frac{n}{n+1})^b=a$

$L=\lim_{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L>1 the series diverges, if L<1 the series converges absolutely, and if L=1 the series may either converge or diverge.

Since $a<\frac{1}{999}<1$ the ratio test concludes that the series converges absolutely.

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Example Question #13 : Ratio Test

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

$\sum_{n = 1}^{\infty} \frac{6}{5^n}$

Note: $f(n) = \frac{6}{5^n}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0.2$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 10$

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

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

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

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0.2$

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 $\lim_{n\rightarrow \infty}$ .

$\frac{\frac{1}{5^{n+1}}}{\frac{1}{5^n}}$ $= \frac{5^n}{5^{n+1}}=0.2$

$\lim_{n\rightarrow \infty} 0.2 \rightarrow 0.2$

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Example Question #14 : Ratio Test

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

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

Note: $f(n) = \frac{10}{2^n}$

Possible Answers:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 10$

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

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

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

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0.5$

Correct answer:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0.5$

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 $\lim_{n\rightarrow \infty}$

$\frac{\frac{1}{2^{n+1}}}{\frac{1}{2^n}}$ $= \frac{2^n}{2^{n+1}}=0.5$

$= \lim_{n\rightarrow \infty} 0.5 \rightarrow 0.5$

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Example Question #11 : Ratio Test

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

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

Note: $f(n) = \frac{e^n}{3^n}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 $\lim_{n\rightarrow \infty}$ .

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

$\lim_{n\rightarrow \infty} \frac{e}{3} \rightarrow \frac{e}{3}$

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Example Question #16 : Ratio Test

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

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

Note: $f(n) = \frac{3^n}{2^n}$

Possible Answers:

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

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

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

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0.5$

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1.5$

Correct answer:

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 1.5$

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 $\lim_{n\rightarrow \infty}$ .

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

$\frac{3}{2}= 1.5$

$=\lim_{n\rightarrow \infty} 1.5 \rightarrow 1.5$

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Example Question #17 : Ratio Test

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

$\sum_{n = 1}^{\infty} \frac{e^n}{5^n}$

Note: $f(n) = \frac{e^n}{5^n}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 $\lim_{n\rightarrow \infty}$ .

$\frac{\frac{e^{n+1}}{5^{n+1}}}{\frac{e^n}{5^n}}$ $= \frac{\frac{e^{n+1}}{e^n}}{\frac{5^{n+1}}{5^n}}$ $= \frac{e}^{5}$

$\lim_{n\rightarrow \infty} \frac{e}{5} \rightarrow \frac{e}{5}$

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Example Question #18 : Ratio Test

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

$\sum_{n = 1}^{\infty} \frac{10e^n}{4^n}$

Note: $f(n) = \frac{e^n}{4^n}$

Possible Answers:

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

$Divergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 10\frac{e}{4}$

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

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

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

Correct answer:

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

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 $\lim_{n\rightarrow \infty}$ .

$\frac{\frac{e^{n+1}}{4^{n+1}}}{\frac{e^n}{4^n}}$ $= \frac{\frac{e^{n+1}}{e^n}}{\frac{4^{n+1}}{4^n}}=\frac{e}{4}$ $,\lim_{n\rightarrow \infty} \frac{e}{4} \rightarrow$ $\frac{e}^{4}$

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Example Question #19 : Ratio Test

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

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

Note: $f(n) = \frac{n!}{e^n}$

Possible Answers:

$Convergent: \lim_{n\rightarrow \infty} \frac{f(n+1)}{f(n)} = 0.25$

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

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

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

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

Correct answer:

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

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 $\lim_{n\rightarrow \infty}$ .

$\frac{\frac{(n+1)!}{e^{n+1}}}{\frac{n!}{e^n}}$ $= \frac{\frac{(n+1)!}{n!}}{\frac{e^{n+1}}{e^n}}$ $= \frac{n+1}^{e}$ $,\lim_{n\rightarrow \infty} \frac{n+1}{e} \rightarrow \infty$

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Calculus 2 : Ratio Test

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #41 : Series In Calculus

Example Question #11 : Ratio Test

Example Question #11 : Ratio Test

Example Question #13 : Ratio Test

Example Question #14 : Ratio Test

Example Question #11 : Ratio Test

Example Question #16 : Ratio Test

Example Question #17 : Ratio Test

Example Question #18 : Ratio Test

Example Question #19 : Ratio Test

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