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Calculus 2 : Ratio Test

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Example Question #101 : Ratio Test

Determine the convergence or divergence of the following series:

$\sum_{n=0}^{\infty}\frac{(n+3)!}{10^{n+1}}$

Possible Answers:

The series is divergent

The series is convergent

The series is conditionally convergent

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

Correct answer:

The series is divergent

Explanation:

To determine the convergence or divergence of the following series, we must use the Ratio Test, which states that for the series $\sum_{n=0}^{\infty}a_{n}$ , and $L=\lim_{n\rightarrow \infty}\left | \frac{a_{n+1}}{a_{n}} \right |$ , that when L>1 , the series diverges, when L<1 , the series converges, and when L=1 , the series may be absolutely convergent, conditionally convergent, or divergent.

For our series, using the above formula, we get $L=\lim_{n\rightarrow \infty} \left | \frac{(n+4)!}{10^{n+2}}\cdot \frac{10^{n+1}}{(n+3)!}\right |$

which simplified becomes

$L=\lim_{n\rightarrow \infty} \left | \frac{n+4}{10}\right |=\infty$

Note that the factorial was simplified by rewriting (n+4)!=(n+4)(n+3)! .

In the limit, the n in the numerator goes to infinity, which makes L go to infinity, which is greater than 1. Thus, the series diverges.

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Example Question #102 : Ratio Test

Determine if the series converges or diverges:

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

Possible Answers:

The series diverges

The series converges

The series may absolutely converge, conditionally converge, or diverge

The series conditionally converges

Correct answer:

The series converges

Explanation:

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

Now, we must find L:

$L=\lim_{n\rightarrow \infty} \left | \frac{2^{n+3}}{(n+3)!} \cdot \frac{(n+2)!}{2^{n+2}}\right |=\lim_{n\rightarrow \infty}\left | \frac{2}{n+3} \right |=0$

We simplified the limit using the properties of radicals and exponents. The denominator of the limit becomes infinitely large, so the term goes to zero. L is therefore less than 1, so the series converges.

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Example Question #103 : Ratio Test

Determine if the series converges (absolutely or conditionally) or diverges:

$\sum_{n=0}^{\infty}\frac{n^2+1}{5^n}$

Possible Answers:

Conditionally Converges

Absolutely Converges

May Absolutely or Conditionally converge, or diverge.

Diverges

Correct answer:

Absolutely Converges

Explanation:

Using the ratio test $\lim_{n \to \infty}\frac{a_{n+1}}{a_n}$ , we get $\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2+1}{5^{n+1}}*\frac{5^n}{n^2+1}\right |=\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2+1}{5n^2+5} \right |$ . Evaluating the limit, we get $\frac{1}{5}$ . Because $\frac{1}{5}<1$ the series converges. Also, because no n value could make the series negative, it absolutely converges .

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Example Question #101 : Ratio Test

Determine whether the series converges or diverges:

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

Possible Answers:

Conditionally Converges

May converge or diverge

Absolutely Converges

Diverges

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 #141 : Series In Calculus

Does the series converge?

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

Possible Answers:

Converges

Does not converge

Converges conditionally

Converges in an interval

Correct answer:

Converges

Explanation:

Untitled

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Example Question #101 : Ratio Test

Does the following series converge or diverge?

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

Possible Answers:

Converge

Diverge

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 #107 : Ratio Test

Does the following series converge or diverge?

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

Possible Answers:

Conditionally converge

Diverge

Absolutely converge

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 #108 : 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 #101 : Ratio Test

Find the interval of convergence of the following series

Possible Answers:

(-1,1)

(-2,1)

[-1,1)

[-1,1]

Correct answer:

[-1,1)

Explanation:

Untitled

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Calculus 2 : Ratio Test

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #101 : Ratio Test

Example Question #102 : Ratio Test

Example Question #103 : Ratio Test

Example Question #101 : Ratio Test

Example Question #141 : Series In Calculus

Example Question #101 : Ratio Test

Example Question #107 : Ratio Test

Example Question #108 : Ratio Test

Example Question #109 : Ratio Test

Example Question #101 : Ratio Test

All Calculus 2 Resources

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