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AP Calculus BC : AP Calculus BC

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Example Questions

Example Question #181 : Calculus

Which of these series cannot be tested for convergence/divergence properly using the ratio test? (Which of these series fails the ratio test?)

Possible Answers:

$\sum_{k=0}^{\infty} \frac{k^2}{3^k}$

None of the other answers.

$\sum_{k=0}^{\infty} \frac{1}{k!}$

$\sum_{k=0}^{\infty} \frac{(-2)^k}{k!}$

$\sum_{k=0}^{\infty} k$

Correct answer:

$\sum_{k=0}^{\infty} k$

Explanation:

The ratio test fails when $\lim_{k\to \infty} \left |\frac{a_{k+1}}{a_k} \right | =1$ . Otherwise the series converges absolutely if $\lim_{k\to \infty} \left |\frac{a_{k+1}}{a_k} \right | <1$ , and diverges if $\lim_{k\to \infty} \left |\frac{a_{k+1}}{a_k} \right | >1$ .

Testing the series $\sum_{k=0}^{\infty} k$ , we have

$\lim_{k\to\infty} \left |\frac{a_{k+1}}{a_k} \right | = \lim_{k\to\infty} \left |\frac{(k+1)}{(k)} \right | = \lim_{k\to\infty} 1+ \frac{1}{k} = 1.$

Hence the ratio test fails here. (It is likely obvious to the reader that this series diverges already. However, we must remember that all intuition in mathematics requires rigorous justification. We are attempting that here.)

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Example Question #182 : Calculus

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:

We cannot conclude when we use the ratio test.

It is convergent.

$-\infty$

$\infty$

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.

$\infty$

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

$\frac{4}{5}$

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 #1 : Ratio Test

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:

We can't conclude when using the ratio test.

$\frac{3}{211}$

The series is convergent.

$\frac{\pi}{e}$

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 #21 : Ratio Test And Comparing Series

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 divergent.

The series is convergent.

We can't use the ratio test here.

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 #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 converges.

The series diverges.

Unable to determine

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:

Unable to determine

The series converges.

The series diverges.

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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Example Question #1 : Riemann Sum: Left Evaluation

Find the Left Riemann sum of the function

f(x)=x+7

on the interval [0,8] divided into four sub-intervals.

Possible Answers:

100

Correct answer:

Explanation:

The interval [0, 8] divided into four sub-intervals gives rectangles with vertices of the bases at

x=0,2,4,6,8

For the Left Riemann sum, we need to find the rectangle heights which values come from the left-most function value of each sub-interval, or f(0), f(2), f(4), and f(6).

f(0) = 0+7 = 7

f(2) = 2+7 = 9

f(4) = 4+7 = 11

f(6) = 6+7 = 13

Because each sub-interval has a width of 2, the Left Riemann sum is

A = bh = 2(7+9+11+13) = 80

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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 #181 : Calculus

Example Question #182 : Calculus

Example Question #2 : Ratio Test

Example Question #1 : Ratio Test

Example Question #21 : Ratio Test And Comparing Series

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

Example Question #1 : Riemann Sum: Left Evaluation

All AP Calculus BC Resources

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