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AP Calculus BC : Ratio Test and Comparing Series

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Example Questions

Example Question #2 : Ratio Test

Assuming that $a_{n}=n^{\alpha}(n+1)$ , $\alpha \ge 2$ . Using the ratio test, what can we say about the series:

$\sum_{n=1}^{\infty} \frac{1}{a_{n}}$

Possible Answers:

$\infty$

$-\infty$

We cannot conclude when we use the ratio test.

It is convergent.

Correct answer:

We cannot conclude when we use the ratio test.

Explanation:

As required by this question we will have to use the ratio test. $L=\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series could either converge or diverge.

To do so, we will need to compute : $\frac{a_{n+1}}{a_{n}}$ . In our case:

$a_{n}=\frac{1}{a_{n}}$

Therefore

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

We know that $\lim_{n\rightarrow \infty }\frac{n}{n+1}=1$

This means that

$\frac{a_{n+1}}{a_{n}}=(\frac{n}{n+1})^{\alpha}\frac{n+1}{n+2}=1\forall\quad \alpha$

Since L=1 by the ratio test, we can't conclude about the convergence of the series.

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Example Question #2 : Ratio Test

Using the ratio test,

what can we say about the series.

$\sum_{_n=1}^{\infty} \frac{1}{n^p}$ where is an integer that satisfies:

p> 1

Possible Answers:

We can't conclude when we use the ratio test.

$\frac{4}{5}$

We can't use the ratio test to study this series.

$\infty$

Correct answer:

We can't conclude when we use the ratio test.

Explanation:

Let $a_{n}$ be the general term of the series. We will use the ratio test to check the convergence of the series.

The Ratio Test states:

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

then if,

1) L<1 the series converges absolutely.

2) L>1 the series diverges.

3) L=1 the series either converges or diverges.

Therefore we need to evaluate,

$\frac{a_{n+1}}{a_{n}}$

we have,

$a_{n+1}=\frac{1}{(n+1)^p} ,a_{n}=\frac{1}{n^p}$

therefore:

$\frac{a_{n+1}}{a_{n}}=\frac{n^p}{(n+1)^p}$ .

We know that

$\lim_{n\rightarrow \infty}\frac{n}{n+1}=1$

and therefore,

$\lim_{n\rightarrow \infty}(\frac{n}{n+1})^p=1$

This means that :

$\lim_{n\rightarrow \infty}\frac{a_{n}}{a_{n+1}}=1$

By the ratio test we can't conclude about the nature of the series. We will have to use another test.

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Example Question #41 : Series Of Constants

Consider the following series :

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

$a_{n}=\frac{n}{n^\alpha +1},\alpha \ge 3$ . Using the ratio test, find the nature of the series.

Possible Answers:

The series is convergent.

$\frac{3}{211}$

We can't conclude when using the ratio test.

$\frac{\pi}{e}$

Correct answer:

We can't conclude when using the ratio test.

Explanation:

Let $a_{n}$ be the general term of the series. We will use the ratio test to check the convergence of the series.

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

We need to evaluate,

$\frac{a_{n+1}}{a_{n}}$ we have:

$a_{n+1}= \frac{n+1}{(n+1)^\alpha +1},a_{n}=\frac{n}{n^\alpha +1}$ .

Therefore:

$\frac{a_{n+1}}{a_{n}}=\frac{n^\alpha +1}{(n+1)^\alpha +1}\frac{n+1}{n}$ . We know that,

$\lim_{n\rightarrow \infty}\frac{n}{n+1}=1$ and therefore

$\lim_{n\rightarrow \infty}\frac{n^\alpha +1}{(n+1)^\alpha +1}=1$

This means that :

$\lim_{n\rightarrow \infty}\frac{a_{n}}{a_{n+1}}=1$ .

By the ratio test we can't conclude about the nature of the series. We will have to use another test.

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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:

The series is convergent.

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

We can't use the ratio test here.

The series is divergent.

We will need to know the values of to decide.

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 #21 : Ratio Test And Comparing Series

Use the ratio test to determine if the series diverges or converges:

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

Possible Answers:

The series converges.

Unable to determine.

The series diverges.

Correct answer:

The series diverges.

Explanation:

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

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

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

This limit is infinite, so the series diverges.

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Example Question #22 : Ratio Test And Comparing Series

Use the ratio test to determine if this series diverges or converges:

$\sum_{n=0}^{\infty} \frac{n^2}{10^n}$

Possible Answers:

The series diverges

The series converges

Unable to determine

Correct answer:

The series converges

Explanation:

$\lim_{n \to \infty} \left | \frac{(n+1)^2}{10^{n+1}} \cdot \frac{ 10^n}{ n^2 }\right |$

$\lim_{n \to \infty} \left | \frac{(n+1)^2}{10} \cdot \frac{ 1}{ n^2 }\right |$

$\lim_{n \to \infty} \left | \frac{n^2+2n+1}{10n^2} \right | = \frac{1}{10}$

Since the limit is less than 1, the series converges.

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Example Question #23 : Ratio Test And Comparing Series

Use the ratio test to determine if the series $\sum_{n=1}^{ \infty } e^{2n+1}$ converges or diverges.

Possible Answers:

The series diverges.

Unable to determine

The series converges.

Correct answer:

The series diverges.

Explanation:

$\lim_{n \to \infty} \left | \frac{e^{2n+3}}{e^{2n+1}}\right | = \lim_{n \to \infty} \left | e^2 \right | = e^2 > 1$

The series diverges.

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Example Question #24 : Ratio Test And Comparing Series

Use the ratio test to determine if the series diverges or converges:

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

Possible Answers:

The series converges.

The series diverges.

Unable to determine

Correct answer:

The series converges.

Explanation:

$\lim_{n \to \infty } \left | \frac{(-3)^{n+1}}{(n+1)!} \cdot \frac{ n!}{(-3)^n}\right |$

$\lim_{n \to \infty } \left | \frac{-3 \cdot(-3)^{n}}{(n+1)\cdot n!} \cdot \frac{ n!}{(-3)^n}\right |$

$\lim_{n \to \infty } \left | \frac{-3}{n+1}\right | = 0 < 1$

The series converges.

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AP Calculus BC : Ratio Test and Comparing Series

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #2 : Ratio Test

Example Question #2 : Ratio Test

Example Question #41 : Series Of Constants

Example Question #11 : Ratio Test

Example Question #21 : Ratio Test And Comparing Series

Example Question #22 : Ratio Test And Comparing Series

Example Question #23 : Ratio Test And Comparing Series

Example Question #24 : Ratio Test And Comparing Series

All AP Calculus BC Resources

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