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Calculus 2 : Series in Calculus

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

Example Questions

Example Question #102 : Convergence And Divergence

Determine whether the series converges or diverges:

$\sum_{n=0}^{\infty}3^nn!$

Possible Answers:

May converge or diverge

Diverges

Absolutely Converges

Conditionally Converges

Correct answer:

Diverges

Explanation:

Using the ratio test, we get $\lim_{n\rightarrow \infty}\left | \frac{3^{n+1}(n+1)!}{3^nn!} \right |=\lim_{n\rightarrow \infty}\left | 3(n+1)\right |=\infty$ and therefore, the series diverges. We simplify the inside by saying $3^{n+1}=3^n*3^1$ and that (n+1)!=(n+1)n! which cancels out with the denominator.

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Example Question #103 : Convergence And Divergence

Does the series converge?

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

Possible Answers:

Does not converge

Converges

Converges in an interval

Converges conditionally

Correct answer:

Converges

Explanation:

Untitled

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Example Question #103 : Convergence And Divergence

Does the following series converge or diverge?

$\sum_{n=0}^{\infty}3^nn!$

Possible Answers:

Diverge

Converge

Correct answer:

Diverge

Explanation:

To determine the convergence or divergence of the series, the most apparent way to do so is by using the ratio test. The ratio test states that a series will converge if:

$\lim_{n \to \infty}\left | \frac{a_{n+1}}{a_{n}} \right |=L<1$

The series will diverge if:

$\lim_{n \to \infty}\left | \frac{a_{n+1}}{a_{n}} \right |=L> 1$

The ratio test can be used by first writing all n as n+1 in the numerator and the denominator is the normal series:

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

This simplifies to:

$\lim_{n \to \infty}\left | 3n+1\right |=\infty$

Because the limit is larger than one, the series diverges.

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Example Question #101 : Ratio Test

Does the following series converge or diverge?

$\sum_{n=0}^{\infty}\frac{(-9)^n}{3^{n+2}(n-1)}$

Possible Answers:

Absolutely converge

Conditionally converge

Diverge

The series either diverges, absolutely converges, or conditionally converges.

Correct answer:

Diverge

Explanation:

To determine the convergence or divergence of the series, the most apparent way to do so is by using the ratio test. The ratio test states that a series will converge if:

$\lim_{n \to \infty}\left | \frac{a_{n+1}}{a_{n}} \right |=L<1$

The series will diverge if:

$\lim_{n \to \infty}\left | \frac{a_{n+1}}{a_{n}} \right |=L> 1$

The ratio test can be used by first writing all n as n+1 in the numerator and the denominator is the normal series:

$\sum_{n=0}^{\infty}\left |\frac{(-9)^{n+1}}{3^{n+3}(n)}\cdot \frac{3^{n+2}(n-1)}{(-9)^n} \right |$

This simplifies to:

$\lim_{n \to \infty}\left | \frac{-9(n-1)}{3n}\right |=3$

Because the limit is larger than one, the series diverges.

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Example Question #105 : Ratio Test

Use limit ratio test for positive series to determine if the following series diverges or converges:

$\sum_{n=1}^{\infty}\frac{n+6}{n^4+5n-4}$

Possible Answers:

Converges

Diverges

Correct answer:

Converges

Explanation:

Consider the following limit:

$\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{n+6}{n^4+5n-4}=\frac{n}{n^4}=\frac{1}{n^3}\:$

Therefore,

$as\: n\rightarrow \infty,\sum_{n=1}^{\infty}\frac{n+6}{n^4+5n-4}=\sum_{n=1}^{\infty}\frac{1}{n^3}=\sum_{n=1}^{\infty}b_n$

Then, according to limit ratio test for positive series:

$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{\frac{n+6}{n^4+5n-4}}{\frac{1}{n^3}}=1$

K=1, therefore both $a_n\: and\: b_n$ converge or diverge.

We know, that $\sum_{n=1}^{\infty}\frac{1}{n^3}$ is generalized harmonic series with p=3, therefore it converges. Consecutively, by limit ratio test, $\sum_{n=1}^{\infty}\frac{n+6}{n^4+5n-4}$ also converges.

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Example Question #109 : Ratio Test

Compute the limit $L =\lim_{n->\infty} \left | \frac{a_{n+1}}{a_n}\right |$ to determine if the series

$\sum_{i=n}^{\infty}a_n$ converges or diverges using the ratio test.

$a_n = \frac{(-1)^{2n+1}(2n+1)}{3^{2n+1}}$

Possible Answers:

The series is absolutely convergent, and therefore converges.

L = 2/3

The series diverges.

L = 2/3

The series diverges.

L = 1/9

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

L = 1

The series is absolutely convergent, and therefore converges.

L = 1/9

Correct answer:

The series is absolutely convergent, and therefore converges.

L = 1/9

Explanation:

Compute the limit $L =\lim_{n->\infty} \left | \frac{a_{n+1}}{a_n}\right |$ to determine if the series

$\sum_{i=n}^{\infty}a_n$ converges or diverges, or neither.

$a_n = \frac{(-1)^{2n+1}(2n+1)}{3^{2n+1}}$

______________________________________________________________

Define for the sequence a_n the limit $L = \lim_{n->\infty}\left | \frac{a_{n+1}}{a_n}\right |$ . The ratio test states:

the series $\sum_{i=1}^{n}a_n$

If L<1 then the series $\sum_{i=1}^{n}a_n$ converges absolutely, and therefore converges.

If L>1 then the series $\sum_{i=1}^{n}a_n$ diverges.

If L = 1 then the series $\sum_{i=1}^{n}a_n$ either converges absolutely, is divergent, or conditionally convergent.

________________________________________________________________

To write the numerator in the limit we need to compute, note that every 2n+1 will become 2(n+1)+1 which simplifies down to 2n+3 . So we have for $a_{n+1}$ ,

$a_{n+1} = \frac{(-1)^{2n+3}(2n+3)}{3^{2n+3}}$

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

$\frac{a_{n+1}}{a_n } = \frac{(-1)^{2n+3}(2n+3)}{3^{2n+3}}\times\frac{3^{2n+1}}{(-1)^{2n+1}(2n+1)}$

$\frac{a_{n+1}}{a_n } = \frac{(-1)^{2n}(-1)^3(2n+3)}{3^{2n}\times 3^3}\times\frac{3^{2n}\times 3 }{(-1)^{2n}(-1)(2n+1)}$

$\frac{a_{n+1}}{a_n } = \frac{(2n+3)}{3^3}\times\frac{3 }{(2n+1)}$

$\frac{a_{n+1}}{a_n }=\frac{2n+3}{9(2n+1)}=\frac{2n+3}{18n+9}$

$L = \lim_{n->\infty}\left |\frac{2n+3}{18n+9} \right | =\lim_{n->\infty}\frac{2n+3}{18n+9}=\lim_{n->\infty}\frac{2+\frac{3}{n}}{18+\frac{9}{n}}=\frac{1}{9}$

$L = \frac{1}{9}$

Because L<1 we can conclude that the series $\sum_{i=n}^{\infty}a_n$ converges by the ratio test.

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Example Question #104 : Convergence And Divergence

Find the interval of convergence of the following series

Possible Answers:

[-1,1]

(-1,1)

[-1,1)

(-2,1)

Correct answer:

[-1,1)

Explanation:

Untitled

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Example Question #111 : Ratio Test

Find the interval of convergence of the following series

Possible Answers:

[4,6)

(4,6)

[4,6]

(4,6)

Correct answer:

(4,6)

Explanation:

Untitled

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Example Question #111 : Ratio Test

Determine if the series converges or diverges:

$\sum_{n=0}^{\infty} \frac{4^{n+2}(n!)}{3^n}$

Possible Answers:

The series may absolutely converge, conditionally converge, or diverge

The series converges

The series diverges

The series conditionally converges

Correct answer:

The series diverges

Explanation:

To determine the convergence of the series, we must use the ratio test, which states that for the series $\sum_{n=0}^{\infty} a_n$ , and $L=\lim_{n\rightarrow \infty} \left | \frac{a_{n+1}}{a_n} \right |$ , if L is greater than 1, the series diverges, if L is equal to 1, the series may absolutely converge, conditionally converge, or diverge, and if L is less than 1 the series is (absolutely) convergent.

For our series, we get

$L=\lim_{n\rightarrow \infty} \left | \frac{4^{n+3}(n+1)!}{3^{n+1}}\cdot \frac{ 3^n}{n!(4^{n+2})}\right |$

Using the properties of radicals and exponents to simplify, we get

$\lim_{n\rightarrow \infty}\frac{4(n+1)}{3}=\infty>1$

L is greater than 1 so the series is divergent.

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Example Question #111 : Ratio Test

Determine whether the series converges or diverges:

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

Possible Answers:

The series conditionally converges

The series converges

The series may absolutely converge, conditionally converge, or diverge

The series diverges

Correct answer:

The series converges

Explanation:

For our series, we get

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

Using the properties of radicals and exponents to simplify, we get

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

L is less than 1 so the series is convergent.

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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 #102 : Convergence And Divergence

Example Question #103 : Convergence And Divergence

Example Question #103 : Convergence And Divergence

Example Question #101 : Ratio Test

Example Question #105 : Ratio Test

Example Question #109 : Ratio Test

Example Question #104 : Convergence And Divergence

Example Question #111 : Ratio Test

Example Question #111 : Ratio Test

Example Question #111 : Ratio Test

All Calculus 2 Resources

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