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Calculus 2 : Ratio Test

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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)

Correct answer:

(4,6)

Explanation:

Untitled

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Example Question #112 : Ratio Test

Determine if the series converges or diverges:

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

Possible Answers:

The series converges

The series conditionally converges

The series diverges

The series may absolutely converge, conditionally converge, or diverge

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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Example Question #114 : Ratio Test

Use the ratio test on the given series and interpret the result.

$\small \sum_{n=0}^{\infty}\frac{(-1)^{2n+1}(3n-2)}{4^{2n+1}}$

Possible Answers:

The series is conditionally convergent.

The series is absolutely convergent, and therefore convergent

The series is divergent

The series is either divergent, conditionally convergent, or absolutely convergent.

The series is convergent, but not absolutely convergent.

Correct answer:

The series is absolutely convergent, and therefore convergent

Explanation:

$\small \sum_{n=0}^{\infty}\frac{(-1)^{2n+1}(3n-2)}{4^{2n+1}}$

_____________________________________________________________

The ratio test can be used to prove that an infinite series $\small \sum_{n=0}^{\infty}a_n$ is convergent or divergent. In some cases, however, the ratio test may be inconclusive.

Define:

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

If...

$\small L <1$ then the series is absolutely convergent, and therefore convergent.

$\small L>1$ then the series is divergent,

$\small L=1$ then the series is either conditionally convergent, absolutely convergent, or divergent.

______________________________________________________________

To compute the limit, we first need to write the expression for $\small a_{n+1}$ .

$\small \small a_{n+1}=\frac{(-1)^{2(n+1)+1}[3(n+1)-2]}{4^{2(n+1)+1}}$

$\small \small \small a_{n+1}=\frac{(-1)^{2n+3}(3n+1)}{4^{2n+3}}$

Now we can find the limit,

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

$\small L=\small \small \lim_{n->0}\left |\frac{ (-1)^{2n+3}(3n+1)}{4^{2n+3}}\times \frac{4^{2n+1}}{(-1)^{2n+1}(3n-2)}\right |$

We can cancel all factors with the exponent.

$\small L=\small \small \lim_{n->0}\left |\frac{ (-1)^{3}(3n+1)}{4^{3}}\times \frac{4^1}{(-1)^1(3n-2)}\right |$

Now the negative factor in the numerator (-1)^3 = -1 will cancel with in the denominator, although this is trivial since we are taking the absolute value regardless.

Continue simplifying,

$L = \lim_{n->0}\left | \frac{3n+1}{16}\times \frac{1}{3n-2}\right |$

$L = \lim_{n->0}\left | \frac{3n+1}{48n-32}\right |=\frac{3}{48}=\frac{1}{16}$

Therefore,

L<1

This proves that the series $\small \sum_{n=0}^{\infty}\frac{(-1)^{2n+1}(3n-2)}{4^{2n+1}}$ is absolutely convergent, and therefore convergent.

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Example Question #115 : Ratio Test

For the following series, perform the ratio test and interpret the results.

$\small \sum_{n=0}^{\infty}(-1)^n(6n^2+1)$

Possible Answers:

Divergent

The series is either conditionally convergent, absolutely convergent, or divergent.

Absolutely Convergent

Conditionally Convergent

Correct answer:

The series is either conditionally convergent, absolutely convergent, or divergent.

Explanation:

$\small \sum_{n=0}^{\infty}(-1)^n(6n^2+1)$

______________________________________________________________

The ratio test can be used to prove that an infinite series $\small \sum_{n=0}^{\infty}a_n$ is convergent or divergent. In some cases, however, the ratio test may be inconclusive.

Define:

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

If...

$\small L <1$ then the series is absolutely convergent, and therefore convergent.

$\small L>1$ then the series is divergent,

$\small L=1$ then the series is either conditionally convergent, absolutely convergent, or divergent.

______________________________________________________________

$L=\lim_{n->\infty}\left |\frac{a_{n+1}}{a_n} \right |= \lim_{n->\infty}\left |\frac{ (-1)^{n+1}[6(n+1)^2+1]}{(-1)^n(6n^2+1)} \right |$

$=\lim_{n->\infty}\left | \frac{(-1) (6n^2+12n+7)}{6n^2+1} \right|$

$=\lim_{n->\infty} \frac{6n^2+12n+7}{6n^2+1}$

Divide above and below by n^2 , in the limit the all terms disappear except for the coefficients on the leading terms. The limit is therefore equal to $\frac{6}{6}=1$

L = 1

Therefore, the series is either conditionally convergent, absolutely convergent, or divergent.

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Calculus 2 : Ratio Test

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

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Example Question #111 : Ratio Test

Example Question #112 : Ratio Test

Example Question #111 : Ratio Test

Example Question #114 : Ratio Test

Example Question #115 : Ratio Test

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