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Calculus 2 : Series in Calculus

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Example Questions

Example Question #1 : Ratio Test

Using the ratio test,

what can we say about the series.

$\sum_{_n=1}^{\infty} \frac{1}{n^p}$ where is an integer that satisfies:

p> 1

Possible Answers:

We can't conclude when we use the ratio test.

$\frac{4}{5}$

We can't use the ratio test to study this series.

$\infty$

Correct answer:

We can't conclude when we use the ratio test.

Explanation:

Let $a_{n}$ be the general term of the series. We will use the ratio test to check the convergence of the series.

The Ratio Test states:

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

then if,

1) L<1 the series converges absolutely.

2) L>1 the series diverges.

3) L=1 the series either converges or diverges.

Therefore we need to evaluate,

$\frac{a_{n+1}}{a_{n}}$

we have,

$a_{n+1}=\frac{1}{(n+1)^p} ,a_{n}=\frac{1}{n^p}$

therefore:

$\frac{a_{n+1}}{a_{n}}=\frac{n^p}{(n+1)^p}$ .

We know that

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

and therefore,

$\lim_{n\rightarrow \infty}(\frac{n}{n+1})^p=1$

This means that :

$\lim_{n\rightarrow \infty}\frac{a_{n}}{a_{n+1}}=1$

By the ratio test we can't conclude about the nature of the series. We will have to use another test.

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Example Question #1 : Convergence And Divergence

Consider the following series :

$\sum_{n=1}^{\infty}a_{n}$ where $a_{n}$ is given by:

$a_{n}=\frac{n}{n^\alpha +1},\alpha \ge 3$ . Using the ratio test, find the nature of the series.

Possible Answers:

We can't conclude when using the ratio test.

The series is convergent.

$\frac{3}{211}$

$\frac{\pi}{e}$

Correct answer:

We can't conclude when using the ratio test.

Explanation:

Let $a_{n}$ be the general term of the series. We will use the ratio test to check the convergence of the series.

$L=\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series could either converge or diverge.

We need to evaluate,

$\frac{a_{n+1}}{a_{n}}$ we have:

$a_{n+1}= \frac{n+1}{(n+1)^\alpha +1},a_{n}=\frac{n}{n^\alpha +1}$ .

Therefore:

$\frac{a_{n+1}}{a_{n}}=\frac{n^\alpha +1}{(n+1)^\alpha +1}\frac{n+1}{n}$ . We know that,

$\lim_{n\rightarrow \infty}\frac{n}{n+1}=1$ and therefore

$\lim_{n\rightarrow \infty}\frac{n^\alpha +1}{(n+1)^\alpha +1}=1$

This means that :

$\lim_{n\rightarrow \infty}\frac{a_{n}}{a_{n+1}}=1$ .

By the ratio test we can't conclude about the nature of the series. We will have to use another test.

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Example Question #1 : Ratio Test

Use the ratio test to determine whether the series below is convergent or divergent.

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

Possible Answers:

The series is convegent.

$\frac{4}{13}$

The series is divergent.

$\frac{6}{11}$

Correct answer:

The series is divergent.

Explanation:

To use the ratio test, we will need to compute the ratio

$\frac{a_{n+1}}{a_n}$ . Then $L=\lim_{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series either converges or diverges.

We have $a_{n+1}=\frac{(n+1)!}{2^{n+1}}, a_n=\frac{n!}{2^n}$ we have then :

$\frac{a_{n+1}}{a_n}=\frac{2^n(n+1)!}{2^{n+1}n!}$ .

Since we can write :

(n+1)!=(n+1)n!

$\frac{a_{n+1}}{a_n}=\frac{n+1}{2}$

Thus $\lim_{n\rightarrow \infty} \frac{a_{n+1}}{a_n}=\lim_{n\rightarrow \infty} \frac{2^n(n+1)!}{2^{n+1}n!}=\infty$ and because $\infty >1$ the series must diverge.

Therefore we conclude that the series is divergent.

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Example Question #1 : Convergence And Divergence

Using the ratio test , what can you say about the following series:

$\sum_{n=1}^{\infty}\frac{sin(n)}{7^n}$

Possible Answers:

The series has two limits.

The series is divergent.

The series is convergent.

$\frac{sin(1)}{3}$

The series will converge and diverge when it gets close to $\infty$ .

Correct answer:

The series is convergent.

Explanation:

We will use the comparison test to conclude about the convergence of this series. To show that the majorant series is convergent we will have to call upon the ratio test.

$L=\lim_{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series either converges or diverges.

We note first that,

$|sin(n)|\le 1 \forall n$ where n is positive integer.

We have $\left|\frac{sin(n)}{7^n}\right|\le\frac{1}{7^n}$ . By the comparison test, if we can show that the series $\sum_{n=1}^{\infty}\frac{1}{7^n}$ is convergent, then by the comparison test, the series $\sum_{n=1}^{\infty}\frac{sin(n)}{7^n}$ is also convergent.

We consider now the series : $\sum_{n=1}^{\infty}\frac{1}{7^n}$ . We have:

$\frac{a_{n+1}}{a_n}=\frac{1}{7}$ and since $\frac{1}{7}< 1$ we conclude that the series is convergent by the ratio test.

This shows that our series is convergent.

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Example Question #2 : Convergence And Divergence

Suppose that a series has positive terms.

If $\frac{a_n}{a_{n+1}}=\frac{2^n}{3^{n+1}}$ what can we say about the convergency of the series.

Possible Answers:

The series is convergent.

We can't conclude.

The series is divergent.

We will need to know the explicit formula for $a_{n}$ .

We will need to know the first two terms.

Correct answer:

The series is divergent.

Explanation:

We are given that the series has positive terms. We know that

$\frac{a_n}{a_{n+1}}=\frac{2^n}{3^{n+1}}$ . This means that $\frac{a_{n+1}}{a_{n}}=3\frac{3^n}{2^n}}$ .

Now we note that $\lim_{n\rightarrow \infty }\frac{3^n}{2^n}=\infty$ .

Then $L=\lim_{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series either converges or diverges.

Since $\infty>1$ the series will diverge.

The ratio test lets us conclude that the series is divergent.

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Example Question #1 : Convergence And Divergence

We will consider the following series :

$\sum_{n=2}^{\infty}\frac{1}{ln(n)^\alpha}$ .

What can you say about the nature of this series using the ratio test? Assume that $\alpha > 0$ .

Possible Answers:

The series is convergent.

We need to know the exact value of $\alpha$ .

The series converges to $\frac{1}{\alpha}$ .

The nature of the series depends on $\alpha$ .

We can't conclude about the nature of the series.

Correct answer:

We can't conclude about the nature of the series.

Explanation:

Note that for $n\ge 2$ and $\alpha>0$ the series is always positive.

To be able to use the ratio test, we will have to compute the ratio:

$\frac{a_{n+1}}{a_{n}}$ . Then find $L=\lim_{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ . If L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series either converges or diverges.

We have $a_{n+1}=\frac{1}{ln(n+1)^\alpha}$ hence :

$\frac{a_{n+1}}{a_{n}}=\frac{ln(n)^\alpha}{ln(n+1)^\alpha}=\left(\frac{ln(n)}{ln(n+1)}\right)^\alpha$

Therefore :

$\lim_{n\rightarrow \infty } \frac{a_{n+1}}{a_{n}}=1$ since $\lim_{n\rightarrow \infty }(\frac{ln(n)}{ln(n+1)})^\alpha=1$

By the ratio test we cannot conclude about the nature of the series.

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Example Question #1 : Convergence And Divergence

We consider the following series:

$\sum_ {n=1}^{\infty}a_{n}$ where $a_{n}=\frac{1}{n\sqrt[n]{3}}$ .

Using the ratio test what can you say about the nature of the series?

Is it convergent or divergent?

Possible Answers:

The series is convergent.

We can't conclude using the ratio test.

$\frac{1}{2}$

$\frac{1}{3}$

The series is divergent.

Correct answer:

We can't conclude using the ratio test.

Explanation:

We will use the ratio test noting first that the series is positive.

We will compute the ratio:

$\frac{a_{n+1}}{a_n}$ . Note that: $a_{n+1}=\frac{1}{(n+1)\sqrt[n+1]{3}}$

Hence : $\frac{a_{n+1}}{a_n}=\frac{n}{n+1}\frac{\sqrt[n]{3}}{\sqrt[n+1]{3}}$

Now we have $\lim_{n\rightarrow \infty} \frac{n}{n+1}=1$ and $\frac{\sqrt[n]{3}}{\sqrt[n+1]{3}}=3^{{\frac{1}{n}}-\frac{1}{n+1}}=3^{\frac{1}{n(n+1)}}$

and we have $\lim_{n\rightarrow \infty}3^{\frac{1}{n(n+1)}}=\lim_{n\rightarrow \infty}e^{\frac{1}{n(n+1)}ln(3))}=1$

$L=\lim_{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series either converges or diverges.

Therefore the ratio test is inconclusive. We will need to use another test .

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Example Question #41 : Series In Calculus

Let the series $\sum_{n=1}^{\infty}a_{n}$ with $a_{n}=\frac{(-1)^{n+1}}{n}$ determine whether the series is convergent or divergent using the ratio test.

Possible Answers:

We can't conclude when using the ratio test.

$\frac{-1}{ln2}$

We cant use the ratio test for this type of series.

The series is divergent.

The series is convergent.

Correct answer:

We can't conclude when using the ratio test.

Explanation:

Note that the series is alternating. To be able to use the ratio test, we will have to compute the ratio:

$\left|\frac{a_{n+1}}{a_n}\ \right|=\frac{(-1)^{n+2}}{n+1}\cdot \frac{n}{(-1)^{n+1}}=\frac{n}{n+1}$

Now we need to see that :

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

Since $\lim_{n\rightarrow \infty }\left|\frac{a_{n+1}}{a_n}\ \right|=1$ we can't conclude by using the ratio test. We will have to call upon another test to show that the series is convergent.

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Example Question #41 : Series In Calculus

Determine if the statement is true or false.

Assume that the series has positive terms. Furthermore suppose that

$\lim _{n\rightarrow \infty}\frac{a_{n+1}}{a_n}=1$ , then all the series of the form $\sum_{n=1}^{\infty}a_n$ are divergent.

Possible Answers:

We can't conclude in general.

The statement above is false.

The series is convergent if it is positive.

The series is divergent.

There is only one series that satisfies that.

Correct answer:

The statement above is false.

Explanation:

To show that the statement above is false, we will consider the following example.

consider the series $\sum_{n=1}^{\infty}a_{n}$ where $a_{n}$ is given by:

$a_{n}=\frac{1}{n(n+1)}$ clearly the series has positive terms. Furthermore, we have

$\lim_{n\rightarrow \infty}\frac{a_{n+1}}{a_n}=\frac{1}{(n+1)(n+2)}{\cdot \frac{n(n+1)}{1}}=\frac{n}{n+2}=1$ , meaning the series can be either convergent or divergent.

However, the series : $\sum_{n=1}^{\infty}a_{n}$ is convergent (use for example the integral test to see that it is convergent).

Therefore, the original statement is false.

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Example Question #41 : Series In Calculus

We consider the series,

$\sum_{n=1}^{\infty}\frac{a^n}{n^b},0<a<\frac{1}{999} ,b>0$ .

Using the ratio test, what can we conclude about the nature of convergence of this series?

Possible Answers:

The series is divergent.

We will need to know the values of to decide.

We can't use the ratio test here.

The series is convergent.

The series converges to $\frac{a}{b}$ .

Correct answer:

The series is convergent.

Explanation:

Note that the series is positive.

As it is required we will use the ratio test to check for the nature of the series.

We have $\frac{a_{n+1}}{a_n}=\frac{a^{n+1}}{(n+1)^b}\frac{n^b}{a^n}=a\left(\frac{n}{n+1}\right)^b$ .

Therefore, $\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=\lim_{n\rightarrow\infty} a(\frac{n}{n+1})b$ $\lim_{n\rightarrow\infty}\frac{n^b}{(n+1)^b}=a(\lim_{n\rightarrow}\frac{n}{n+1})^b=a$

$L=\lim_{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L>1 the series diverges, if L<1 the series converges absolutely, and if L=1 the series may either converge or diverge.

Since $a<\frac{1}{999}<1$ the ratio test concludes that the series converges absolutely.

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Calculus 2 : Series in Calculus

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Ratio Test

Example Question #1 : Convergence And Divergence

Example Question #1 : Ratio Test

Example Question #1 : Convergence And Divergence

Example Question #2 : Convergence And Divergence

Example Question #1 : Convergence And Divergence

Example Question #1 : Convergence And Divergence

Example Question #41 : Series In Calculus

Example Question #41 : Series In Calculus

Example Question #41 : Series In Calculus

All Calculus 2 Resources

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