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Calculus 2 : Concepts of Convergence and Divergence

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Example Questions

Example Question #1 : Concepts Of Convergence And Divergence

One of the following infinite series CONVERGES. Which is it?

Possible Answers:

$\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$

None of the others converge.

$\sum_{k=1}^{\infty} (-1)^k sin(k)$

$\sum_{k=2}^{\infty} \frac{x}{ln(x)}$

$\sum_{k=1}^{\infty} \frac{1}{k}$

Correct answer:

$\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$

Explanation:

$\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$ converges due to the comparison test.

We start with the equation $\frac{1}{k^2} =\frac{1}{k^2}$ . Since $sin(k) \leq 1$ for all values of k, we can multiply both side of the equation by the inequality and get $\frac{sin(k)}{k^2} \leq \frac{1}{k^2}$ for all values of k. Since $\sum_{k=1}^{\infty} \frac{1}{k^2}$ is a convergent p-series with p=2 , $\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$ hence also converges by the comparison test.

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Example Question #12 : Introduction To Series In Calculus

Determine the nature of the following series having the general term:

$\frac{9}{(9n+3)(9n+89)}$

Possible Answers:

$\frac{3}{9}$

$\frac{2}{9}$

$\frac{1}{9}$

The series is convergent.

$\frac{9}{82}$

Correct answer:

The series is convergent.

Explanation:

We will use the Limit Comparison Test to show this result.

We first denote the genera term of the series by:

$u_{n}=\frac{9}{(9n+3)(9n+89)}$ and $v_{n}=\frac{9}{n^2}$ .

We have $\lim_{n\rightarrow\infty}\frac{u_n}{v_{n}}=1$ and the series have the same nature .

We know that

$\sum_{n=1}^{\infty} u_{n}=\sum_{n=1}^{\infty} \frac{9}{n^2}$ is convergent by comparing the integral

$9\int_{1}^{\infty} \frac{1}{x^2}dx$ which we know is convergent.

Therefore by the Limit Comparison Test.

we have $\sum_{n=1}^{\infty} v_{n}=\sum_{n=1}^{\infty} \frac{9}{(9n+3)(9n+89)}$ .

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Example Question #13 : Introduction To Series In Calculus

If $\sum_{k=1}^{\infty} a_{k}$ converges, which of the following statements must be true?

Possible Answers:

The limit of the $n^{th}$ partial sums as approaches infinity is zero.

For some large value of , $a_{k} \geq (a_{k})^{3}$ .

None of the other answers must be true.

The limit of the $n^{th}$ term as approaches infinity is not zero.

Correct answer:

For some large value of , $a_{k} \geq (a_{k})^{3}$ .

Explanation:

If the series converges, then we know the terms must approach zero. At some point, the terms will be less than 1, meaning when you take the third power of the term, it will be less than the original term.

Other answers are not true for a convergent series by the $n^{th}$ term test for divergence.

In addition, the limit of the $n^{th}$ partial sums refers to the value the series converges to. A convergent series need not converge to zero. The alternating harmonic series is a good counter example to this.

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Example Question #53 : Taylor Series

Which of following intervals of convergence cannot exist?

Possible Answers:

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

$\small [2,10000]$

$\small [2,\infty)$

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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Calculus 2 : Concepts of Convergence and Divergence

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Concepts Of Convergence And Divergence

Example Question #12 : Introduction To Series In Calculus

Example Question #13 : Introduction To Series In Calculus

Example Question #53 : Taylor Series

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

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