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Calculus 2 : Introduction to Series in Calculus

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Example Questions

Example Question #2 : Concepts Of Convergence And Divergence

Which of following intervals of convergence cannot exist?

Possible Answers:

$\small [2,\infty)$

$\small [2,10000]$

For any $\small \epsilon>0$ , the interval $\small [a-\epsilon,a+\epsilon]$ for some $\small a\in\mathbb{R}$

For any $\small q,p$ such that $\small q \leq p$ , the interval [q,p]

Correct answer:

$\small [2,\infty)$

Explanation:

$\small [2,\infty)$ cannot be an interval of convergence because a theorem states that a radius has to be either nonzero and finite, or infinite (which would imply that it has interval of convergence $\small (-\infty,\infty)$ ). Thus, $\small [2,\infty)$ can never be an interval of convergence.

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Example Question #2 : Concepts Of Convergence And Divergence

Which of the following statements is true regarding the following infinite series?

$\small \sum_{n=0}^\infty \sin n$

Possible Answers:

The series diverges, by the divergence test, because the limit of the sequence $\small a_n=\sin n$ does not approach a value as $\small n\to\infty$

The series diverges because $\small \lim_{n\to \infty} \sin n=\alpha$ for some $\small \alpha>0$ and finite.

The series converges because $\small \lim_{n\to\infty} \sin n=0$

The series diverges to $\small \infty$ .

Correct answer:

The series diverges, by the divergence test, because the limit of the sequence $\small a_n=\sin n$ does not approach a value as $\small n\to\infty$

Explanation:

The divergence tests states for a series $\small \sum_{n=0}^\infty a_n$ , if $\small \lim a_n$ is either nonzero or does not exist, then the series diverges.

The limit $\small \small \lim_{n\to\infty} \sin n$ does not exist, so therefore the series diverges.

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Example Question #2 : Concepts Of Convergence And Divergence

Determine whether the following series converges or diverges:

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

Possible Answers:

None of the other answers.

The series converges.

The series diverges.

The series conditionally converges.

Correct answer:

The series converges.

Explanation:

To prove the series converges, the following must be true:

If $\lim_{n\rightarrow \infty }a_{n}$ converges, then $\sum_{n=0}^{\infty }a_{n}$ converges.

Now, we simply evaluate the limit:

$\lim_{n\rightarrow \infty }\frac{n^3+3n}{2n^3}=\frac{1}{2}$

The shortcut that was used to evaluate the limit as n approaches infinity was that the coefficients of the highest powered term in numerator and denominator were divided.

The limit approaches a number (converges), so the series converges.

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Example Question #3 : Concepts Of Convergence And Divergence

Determine whether the following series converges or diverges. If it converges, what does it converge to?

$\sum_{n=0}^{\infty}(2)^{-2n}$

Possible Answers:

$\frac{3}{4}$

Diverges

$\frac{4}{3}$

Correct answer:

$\frac{4}{3}$

Explanation:

First, we reduce the series into a simpler form.

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

We know this series converges because

$\lim_{n \rightarrow \infty}(\frac{1}{4})^{n}=0$

By the Geometric Series Theorem, the sum of this series is given by

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

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Example Question #4 : Concepts Of Convergence And Divergence

Determine whether the following series converges or diverges. If it converges, what does it converge to?

$\sum_{n=1}^{\infty}\frac{1}{n^{-0.5}}$

Possible Answers:

$\frac{1}{1-\sqrt{2}}$

$\frac{\sqrt{2}}{2}$

Diverges

Correct answer:

Diverges

Explanation:

Notice how this series can be rewritten as

$\sum_{n=1}^{\infty}\frac{1}{n^{-0.5}}= \sum_{n=1}^{\infty}\sqrt{n}$

$\lim_{n \rightarrow \infty}\sqrt{n}=\infty$

Therefore this series diverges.

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Example Question #1 : Concepts Of Convergence And Divergence

There are 2 series, $\sum_{n=1}^{\infty}a_{n}$ and $\sum_{n=1}^{\infty}b_{n}$ , and they are both convergent. Is $\sum_{n=1}^{\infty}a_{n} + \sum_{n=1}^{\infty}b_{n}$ convergent, divergent, or inconclusive?

Possible Answers:

Divergent

Inconclusive

Convergent

Correct answer:

Convergent

Explanation:

Infinite series can be added and subtracted with each other.

$\sum_{n=1}^{\infty}a_{n} + \sum_{n=1}^{\infty}b_{n} = \sum_{n=1}^{\infty}( a_{n} + b_{n})$

Since the 2 series are convergent, the sum of the convergent infinite series is also convergent.

Note: The starting value, in this case n=1, must be the same before adding infinite series together.

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Example Question #1 : Concepts Of Convergence And Divergence

You have a divergent series $\sum_{n=1}^{\infty} a_{n}$ , and you multiply it by a constant 10. Is the new series $\sum_{n=1}^{\infty}10* a_{n}$ convergent or divergent?

Possible Answers:

Divergent

Inconclusive

Convergent

Correct answer:

Divergent

Explanation:

This is a fundamental property of series.

For any constant c, if $\sum_{n=1}^{\infty} a_{n}$ is convergent then $\sum_{n=1}^{\infty} c*a_{n}$ is convergent, and if $\sum_{n=1}^{\infty} a_{n}$ is divergent, $\sum_{n=1}^{\infty} c* a_{n}$ is divergent.

$\sum_{n=1}^{\infty} a_{n}$ is divergent in the question, and the constant c is 10 in this case, so $\sum_{n=1}^{\infty} 10*a_{n}$ is also divergent.

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Example Question #21 : Introduction To Series In Calculus

There are 2 series, $\sum_{n=1}^{\infty}a_{n}$ and $\sum_{n=1}^{\infty}b_{n}$ , and they are both divergent. Is $\sum_{n=1}^{\infty}a_{n} + \sum_{n=1}^{\infty}b_{n}$ convergent, divergent, or inconclusive?

Possible Answers:

Inconclusive

Divergent

Convergent

Correct answer:

Inconclusive

Explanation:

$\sum_{n=1}^{\infty}a_{n}$ is divergent

$\sum_{n=1}^{\infty}b_{n}$ is divergent

However, unlike convergent series in which the sum of convergent series will produce a convergent series, this is not the case for divergent series. Due to the nature of infinite series, adding together 2 divergent series may be divergent, but it may also produce a convergent series. More information is needed.

$\sum_{n=1}^{\infty}a_{n} + \sum_{n=1}^{\infty}b_{n}$ is inconclusive

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Example Question #22 : Introduction To Series In Calculus

Use the limit test (divergence test) to find if the series is convergent, divergent, or inconclusive.

$\sum_{n=1}^{\infty}\frac{6^{n+4}}{11^{n}}$

Possible Answers:

Divergent

Inconclusive

Convergent

Correct answer:

Inconclusive

Explanation:

Divergence Test and Limit Test are the same tests with different names.

If $\lim_{n \to \infty} a_{n} \neq 0$ then $a_{n}$ diverges.

However, if $\lim_{n \to \infty} a_{n} = 0$ the test is inconclusive.

Solution:

$\lim_{n \to \infty} a_{n} = \lim_{n \to \infty}\frac{6^{n+4}}{11^{n}}$

$\\=\lim_{n \to \infty} 6*\left(\frac{6}{11}\right)^{n} \\ \\= 6*0 \\ = 0$

The series is inconclusive by the divergence test.

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Example Question #23 : Introduction To Series In Calculus

Use the limit test (divergence test) to find if the series is convergent, divergent, or inconclusive.

$\sum_{n=1}^{\infty}\frac{12+6sin(n)}{n}$

Possible Answers:

Convergent

Divergent

Inconclusive

Correct answer:

Inconclusive

Explanation:

Divergence Test and Limit Test are the same tests with different names.

If $\lim_{n \to \infty} a_{n} \neq 0$ then $a_{n}$ diverges.

However, if $\lim_{n \to \infty} a_{n} = 0$ the test is inconclusive.

Solution:

$\lim_{n \to \infty} a_{n} = \lim_{n \to \infty}\frac{12+6sin(n)}{n}$

$\lim_{n \to \infty} \frac{12}{n} + \lim_{n \to \infty} \frac{6sin(n)}{n} = 0 + 0 = 0$

This series is inconclusive by the divergence test.

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Calculus 2 : Introduction to Series in Calculus

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2 : Concepts Of Convergence And Divergence

Example Question #2 : Concepts Of Convergence And Divergence

Example Question #2 : Concepts Of Convergence And Divergence

Example Question #3 : Concepts Of Convergence And Divergence

Example Question #4 : Concepts Of Convergence And Divergence

Example Question #1 : Concepts Of Convergence And Divergence

Example Question #1 : Concepts Of Convergence And Divergence

Example Question #21 : Introduction To Series In Calculus

Example Question #22 : Introduction To Series In Calculus

Example Question #23 : Introduction To Series In Calculus

All Calculus 2 Resources

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