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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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

Example Questions

Example Question #2911 : Calculus Ii

Find the radius of convergence of

$\sum_{n=0}^{\infty} \frac{x^{5n}}{3^n}$ .

Possible Answers:

$\infty$

$\frac{1}{3^{\frac{1}{5}}}$

$5^{\frac{1}{3}}$

$3^{\frac{1}{5}}$

Correct answer:

$3^{\frac{1}{5}}$

Explanation:

We can find the radius of convergence of

$\sum_{n=0}^{\infty} \frac{x^{5n}}{3^n}$

using the ratio test:

$\lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n} \right |=\lim_{n\to \infty} \left|\frac{x^{5n+5}}{3^{n+1}}\frac{3^n}{x^{5n}} \right |=\lim_{n\to \infty} \left|\frac{x^{5}}{3} \right |=\left|\frac{x^5}{3}\right|<1$

which means that $|x|<3^{\frac{1}{5}}$

So that $R=3^{\frac{1}{5}}$ is the radius of convergence.

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Example Question #92 : Ratio Test

Determine whether the series is convergent or divergent:

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

Possible Answers:

The series may be (absolutely) convergent, divergent, or conditionally convergent

The series is divergent

The series is (absolutely) convergent

The series is conditionally convergent

Correct answer:

The series is divergent

Explanation:

To determine whether the series given converges or diverges, we must use the Ratio Test, which states that for

$L=\lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_{n}}\right |$ , if L<1 , then series is (absolutely) convergent; if L=1 , then the series may be (absolutely) convergent, divergent, or conditionally convergent; and if L>1 , then the series is divergent.

So, we must evaluate the limit as given by the above formula:

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

Thus, the series is divergent.

(The limit was solved by $\lim_{n\rightarrow \infty } \frac{n}{n}(\frac{1+n^{-1}}{3n^{-1}})=\infty$ . As approaches infinity, the negative terms go to zero, which makes zero in the denominator and in the numerator, equalling infinity.)

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Example Question #93 : Ratio Test

Determine if the series converges

$\sum_{n=0}^{\infty }\frac{3^n}{2^{2n+1}}$

Possible Answers:

Series diverges

Cannot be determined

Series converges

Correct answer:

Series converges

Explanation:

In order to test the convergence of the series, we use the ratio test.

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

If...

L<1 the series is absolutely convergent

L=1 the convergence is undetermined

L>1 the series diverges

For this problem

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

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

$=\lim_{n\rightarrow \infty }\left | \frac{3\cdot3^{n}}{2^2\cdot2^{2n+1} } \cdot \frac{2^{2n+1}}{3^n} \right |$

$=\lim_{n\rightarrow \infty }\left | \frac{3}{2^2} } \right |$

$=\frac{3}{4}$

And since

L<1 the series absolutely converges.

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Example Question #2911 : Calculus Ii

Use the ratio test to figure out if the series

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

converges.

Possible Answers:

it diverges to infinity

It converges

Correct answer:

It converges

Explanation:

an easy sequence to compare $\frac{1}{n^2+n}$ to is $\frac{1}{n^2}$ , which we know converges.

$\frac{\frac{1}{n^2}}{\frac{1}{n^2+n}} = \frac{n^2+n}{n^2}$

We need to find $\lim_{n \to \infty} \frac{n^2+n}{n^2} = \lim_{n \to \infty} 1+\frac{1}{n}$

$\frac{1}{n}$ goes to zero so this limit is . Since the limit of the ratio is 1 and $\sum \frac{1}{n^2}$ converges, $\sum \frac{1}{n^2+n}$ must converge also.

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Example Question #92 : Ratio Test

Can we use the Ratio Test to test the convergence/divergence of the infinite series $\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^3}$ ?

Possible Answers:

No, the ratio test fails here.

Yes, and the series converges

Yes, and the series diverges

Correct answer:

No, the ratio test fails here.

Explanation:

To use the ratio test, we must first evaluate

$\lim_{k \to \infty} |\frac{a_{k+1}}{a_k}|$

$=\lim_{k \to \infty} |\frac{\frac{(-1)^{k+1}}{(k+1)^3}}{\frac{(-1)^k}{k^3}}|$

$=\lim_{k \to \infty} |\frac{(-1)^{k+1}}{(k+1)^{3}} \times \frac{k^3}{(-1)^k}|$

$= \lim_{k \to \infty} | -1 \times \frac{k^3}{(k+1)^3}|$

$= \lim_{k \to \infty} \frac{k^3}{(k+1)^3}$

Since the result of the limit is , the Ratio Test fails, and therefore we cannot use it in testing this series for convergence/divergence.

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Example Question #2912 : Calculus Ii

Determine whether the series converges or diverges:

$\sum_{n=0}^{\infty} \frac{(n+2)!}{5^n}$

Possible Answers:

The series converges

The series may be convergent, divergent, or conditionally convergent

The series is conditionally convergent

The series diverges

Correct answer:

The series diverges

Explanation:

To determine whether the series converges or diverges, we must use the ratio test, which states that for the given series $\sum_{n=0}^{\infty} a_{n}$ , and

$L=\lim_{n\rightarrow \infty} \left |\frac{ a_{n+1}}{a_{n}}\right |$ , when L is equal to 1, then the series may be convergent, divergent, or conditionally convergent; when L is less than 1 then the series is convergent; and when L is greater than 1 the series is divergent.

Using the above test, we get

$L=\lim_{n\rightarrow \infty} \left | \frac{(n+3)!}{5^{n+1}} \cdot \frac{5^n}{(n+2)!} \right |=\lim_{n\rightarrow \infty}\left | \frac{n+3}{5^1}\right |=\infty>1$

Because L is greater than 1, the series diverges.

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Example Question #96 : Ratio Test

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

Determine the convergence of the series using the Ratio Test.

Possible Answers:

Series diverges

Cannot be determined

Series converges

Correct answer:

Series converges

Explanation:

The ratio test is defined as follows

$\lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |$

where a_n is the n-th term of the series.

if the limit

is greater than , the series diverges
is less than , the series converges
equal to , the test is inconclusive

Finding the limit for the terms in our series,

$\lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |=\lim_{n \to \infty} \left | \frac{\frac{(n+1)^2}{3^{n+1}}}{\frac{n^2}{3^n}} \right |$

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

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

$=\frac{1}{3}*1$

$=\frac{1}{3}$

Because the limit is less than ,

the series converges.

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Example Question #97 : Ratio Test

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

Determine the convergence of the series using the Ratio Test.

Possible Answers:

Series converges

Cannot be determined

Series diverges

Correct answer:

Series converges

Explanation:

The ratio test is defined as follows

$\lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |$

where a_n is the n-th term of the series.

if the limit

is greater than , the series diverges
is less than , the series converges
equal to , the test is inconclusive

Finding the limit for the terms in our series,

$\lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |=\lim_{n \to \infty} \left | \frac{\frac{(ln\,n+1)3^{n+1}}{2^{2(n+1)+1}}}{\frac{(ln\,n)3^n}{2^(2n+1)}} \right |$

$=\lim_{n \to \infty}\left |\frac{(ln\,n+1)2^{2n+1}*3^{n+1}}{(ln\,n)2^{2n+3}*3^n} \right |$

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

$=\frac{3}{4}*1$

$=\frac{3}{4}$

Because the limit is less than ,

the series converges.

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Example Question #93 : Ratio Test

$\sum_{n=1}^{\infty}\frac{n!}{4^n}$

Possible Answers:

Series converges

Cannot be determined

Series diverges

Correct answer:

Series diverges

Explanation:

The ratio test is defined as follows

$\lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |$

where a_n is the n-th term of the series.

if the limit

is greater than , the series diverges
is less than , the series converges
equal to , the test is inconclusive

Finding the limit for the terms in our series,

$\lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |=\lim_{n \to \infty} \left | \frac{\frac{(n+1)!}{4^{n+1}}}{\frac{n!}{4^n}} \right |$

$=\lim_{n \to \infty}\left |\frac{(n+1)n!4^n}{n!4^{n+1}} \right |$

$=\frac{1}{4}\lim_{n \to \infty}\left |n+1\right |$

$=\frac{1}{4}*\infty$

$=\infty$

Because the limit is greater than ,

the series diverges.

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Example Question #98 : Ratio Test

$\sum_{n=1}^{\infty}\frac{n^3}{n!2^n}$

Determine the convergence of the series.

Possible Answers:

Series converges

Cannot be determined

Series diverges

Correct answer:

Series converges

Explanation:

The ratio test is defined as follows

$\lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |$

where a_n is the n-th term of the series.

if the limit

is greater than , the series diverges
is less than , the series converges
equal to , the test is inconclusive

Finding the limit for the terms in our series,

$\lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right |=\lim_{n \to \infty} \left | \frac{\frac{(n+1)^3}{(n+1)!2^{n+1}}}{\frac{n^3}{n!2^n}} \right |$

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

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

$=\frac{1}{2}*0$

Because the limit is less than ,

the series converges.

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2911 : Calculus Ii

Example Question #92 : Ratio Test

Example Question #93 : Ratio Test

Example Question #2911 : Calculus Ii

Example Question #92 : Ratio Test

Example Question #2912 : Calculus Ii

Example Question #96 : Ratio Test

Example Question #97 : Ratio Test

Example Question #93 : Ratio Test

Example Question #98 : Ratio Test

All Calculus 2 Resources

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