Calculus 2 : Introduction to Series in Calculus

Study concepts, example questions & explanations for Calculus 2

varsity tutors app store varsity tutors android store

Example Questions

Example Question #31 : Introduction To Series In Calculus

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

\(\displaystyle \sum_{n=1}^{\infty}\frac{3n^2+4n+6}{11n^2+7n+4}\)

Possible Answers:

Convergent

Divergent

Inconclusive

Correct answer:

Divergent

Explanation:

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

If \(\displaystyle \lim_{n \to \infty} a_{n} \neq 0\) then \(\displaystyle a_{n}\) diverges.

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

Solution:

\(\displaystyle \lim_{n \to \infty} a_{n} = \lim_{n \to \infty}\frac{3n^2+4n+6}{11n^2+7n+4}\)

\(\displaystyle \lim_{n \to \infty} \frac{3n^2}{11n^2} = \lim_{n \to \infty} \frac{3}{11} = \frac{3}{11}\)

\(\displaystyle \lim_{n \to \infty} a_{n} = \frac{3}{11} \neq 0\)

The series is divergent by the divergence test.

Example Question #32 : Introduction To Series In Calculus

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

\(\displaystyle \sum_{n=1}^{\infty}\frac{ln(n)}{n^3}\)

Possible Answers:

Convergent

Inconclusive

Divergent

Correct answer:

Inconclusive

Explanation:

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

If \(\displaystyle \lim_{n \to \infty} a_{n} \neq 0\) then \(\displaystyle a_{n}\) diverges.

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

Solution:

\(\displaystyle \lim_{n \to \infty} a_{n} = \lim_{n \to \infty}\frac{ln(n)}{n^3}=0\)

Although ln(n) also tends to infinity, n grows to infinity more quickly than ln(n), and so the limit will go to 0.

The series is inconclusive by the divergence test.

Example Question #33 : Introduction To Series In Calculus

Does the following series converge absolutely, conditionally, or is it divergent? 

\(\displaystyle \sum_{n=1}^{\infty}(\frac{-9}{20})^n\)

Possible Answers:

Absolutely Convergent

Conditionally Convergent

Divergent

Correct answer:

Absolutely Convergent

Explanation:

A series is absolutely convergent if \(\displaystyle \sum_{n=1}^{\infty}|a_{n}|\) is convergent.

If \(\displaystyle \sum_{n=1}^{\infty}|a_{n}|\) is not convergent, but \(\displaystyle \sum_{n=1}^{\infty}a_{n}\) is convergent, then it is conditionally convergent.

This nuance matters when testing series with negatives. First test for absolute convergence.

\(\displaystyle \sum_{n=1}^{\infty}|a_{n}|= \sum_{n=1}^{\infty}\left(\frac{9}{20}\right)^n\) is a geometric series, with \(\displaystyle r < 1\).

So this series is absolutely convergent by the geometric test.

Example Question #34 : Introduction To Series In Calculus

Does the following series converge absolutely, conditionally, or is it divergent? 

\(\displaystyle \sum_{n=1}^{\infty}\left(\frac{-15}{2}\right)^n\)

Possible Answers:

Absolutely Convergent

Divergent

Conditionally Convergent

Correct answer:

Divergent

Explanation:

A series is absolutely convergent if \(\displaystyle \sum_{n=1}^{\infty}|a_{n}|\) is convergent.

If \(\displaystyle \sum_{n=1}^{\infty}|a_{n}|\) is not convergent, but \(\displaystyle \sum_{n=1}^{\infty}a_{n}\) is convergent, then it is conditionally convergent.

This nuance matters when testing series with negatives. First test for absolute convergence.

\(\displaystyle a_{n}=\left(\frac{-15}{2}\right)^n\)

\(\displaystyle |a_{n}| = \left(\frac{15}{2}\right)^n\)

\(\displaystyle \sum_{n=1}^{\infty}|a_{n}| = \sum_{n=1}^{\infty}\frac{15}{2}\)

This is a geometric series with r > 1, and |r| > 1, so this series is divergent by the geometric test.

Example Question #35 : Introduction To Series In Calculus

Does the following series converge absolutely, conditionally, or is it divergent? 

\(\displaystyle \sum_{n=1}^{\infty}\frac{(-7)^n}{n^3}\)

Possible Answers:

Divergent

Absolutely Convergent

Conditionally Convergent

Correct answer:

Divergent

Explanation:

A series is absolutely convergent if \(\displaystyle \sum_{n=1}^{\infty}|a_{n}|\) is convergent.

If \(\displaystyle \sum_{n=1}^{\infty}|a_{n}|\) is not convergent, but \(\displaystyle \sum_{n=1}^{\infty}a_{n}\) is convergent, then it is conditionally convergent.

This nuance matters when testing series with negatives. First test for absolute convergence.

\(\displaystyle \sum_{n=1}^{\infty}|a_{n}| = \sum_{n=1}^{\infty} \frac{7^n}{n^3}\)

\(\displaystyle \lim_{n \to \infty}|a_{n}|=\lim_{n \to \infty}\frac{7^n}{n^3}=\infty\)

This series diverges by the divergence test.

Now test for conditional convergence.

\(\displaystyle \lim_{n \to \infty}a_{n} = \lim_{n \to \infty} (-1)^n\frac{7^n}{n^3}\)

This limit bounces between positive and negative infinity, and so the limit does not exist. By the divergence test, this series diverges.

 

Example Question #36 : Introduction To Series In Calculus

Does the following series converge absolutely, conditionally, or is it divergent? 

\(\displaystyle \sum_{n=1}^{\infty}\frac{-e^n}{n^4}\)

Possible Answers:

Conditionally Convergent

Divergent

Absolutely Convergent

Correct answer:

Divergent

Explanation:

A series is absolutely convergent if \(\displaystyle \sum_{n=1}^{\infty}|a_{n}|\) is convergent.

If \(\displaystyle \sum_{n=1}^{\infty}|a_{n}|\) is not convergent, but \(\displaystyle \sum_{n=1}^{\infty}a_{n}\) is convergent, then it is conditionally convergent.

This nuance matters when testing series with negatives. First test for absolute convergence.

\(\displaystyle \lim_{n \to \infty} |a_{n}| = \lim_{n \to \infty}\frac{e^n}{n^4} = \infty\)

The absolute series diverges by the divergence test.

Now test for conditional convergence.

\(\displaystyle a_{n} = (-1)^n\frac{e^n}{n^4}\)

\(\displaystyle \lim_{n \to \infty}a_{n}\) alternates between positive and negative infinity, and so it diverges by the divergence test.

This series diverges.

Example Question #1 : Limits Of Sequences

There are 2 series \(\displaystyle \sum_{n=1}^{\infty} \left(\frac{3}{7}\right)^n\)and \(\displaystyle \sum_{n=1}^{\infty}\left(\frac{5}{11}\right)^n\).

Is the sum of these 2 infinite series convergent, divergent, or inconclusive?

Possible Answers:

Divergent

Inconclusive

Convergent

Correct answer:

Convergent

Explanation:

A way to find out if the sum of the 2 infinite series is convergent or not is to find out whether the individual infinite series are convergent or not.

Test the first series 

\(\displaystyle \sum_{n=1}^{\infty} \left(\frac{3}{7}\right)^n\).

This is a geometric series with \(\displaystyle r = \frac{3}{7} < 1\).

By the geometric test, this series is convergent.

 

Test the second series 

\(\displaystyle \sum_{n=1}^{\infty}\left(\frac{5}{11}\right)^n\).

This is a geometric series with \(\displaystyle r = \frac{5}{11} < 1\).

By the geometric test, this series is convergent.

 

Since both of the series are convergent, \(\displaystyle \sum_{n=1}^{\infty} \left(\frac{3}{7}\right)^n + \sum_{n=1}^{\infty}\left(\frac{5}{11}\right)^n\) is also convergent. 

Learning Tools by Varsity Tutors