P-Series - AP Calculus BC

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Question

Determine if the series converges or diverges. You do not need to find the sum.

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

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Answer

We can compare this to the series $\sum_{n = 1}^{\infty } \frac{1}{n^{3}}$ which we know converges by the p-series test.

To figure this out, let's first compare $n^{3} + 2$ to $n^{3}$ . For any number n, $n^{3} + 2$ will be larger than $n^{3}$ .

There is a rule in math that if you take the reciprocal of each term in an inequality, you are allowed to flip the signs.

Thus, $n^{3} + 2 > n^{3}}$ turns into

$\frac{1}{n^{3}} > \frac{1}{n^{3} + 2}$ .

And so, because $\frac{1}{n ^{3}}$ converges, thus our series also converges.

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Question

True or False, a -series cannot be tested conclusively using the ratio test.

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Answer

We cannot test for convergence of a -series using the ratio test. Observe,

For the series $\sum_{n=1}^\infty \frac{1}{n^p}$ ,

$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} | \frac{n^p}{(n+1)^p}| = \lim_{n \to \infty} (\frac{n}{n+1})^p = 1^p =1$ .

Since this limit is regardless of the value for , the ratio test is inconclusive.

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Question

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

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Answer

$\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$ 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 , $\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$ hence also converges by the comparison test.

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Question

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

$\frac{8}{\sqrt{n(n+1)(n+2)}}$

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Answer

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

We need to note that the following limit

$\frac{\frac{8}{\sqrt{n(n+1)(n+2)}}}{\frac{8}{n^{3/2}}}}$

goes to 1 as n goes to infinity.

Therefore the series have the same nature. They either converge or diverge at the same time.

We will focus on the series:

$\frac{1}{n^{3/2}}$ .

We know that this series is convergent because it is a p-series. (Remember that

$\sum_{n=1}^{\infty}\frac{1}{n^p}$ converges if p>1 and we have p=3/2 which is greater that one in this case)

By the Limit Comparison Test, we deduce that the series is convergent, and that is what we needed to show.

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Question

Does the series $\sum_{n=0}^{\infty }\frac{(-1)^n}{n}$ converge conditionally, absolutely, or diverge?

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Answer

The series converges conditionally.

The absolute values of the series $\sum_{n=0}^{\infty }\left |\frac{(1)^n}{n} \right |$ is a divergent p-series with $p\leq 1$ .

However, the the limit of the sequence $\lim_{n\rightarrow \infty }\frac{(-1)^n}{n}=0$ and it is a decreasing sequence.

Therefore, by the alternating series test, the series converges conditionally.

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Question

Which of the following tests will help determine whether $\sum_{x=1}^{\infty}\frac{1}{x}$ is convergent or divergent, and why?

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Answer

The series $\sum_{x=1}^{\infty}\frac{1}{x}$ is a harmonic series.

The Nth term test and the Divergent test may not be used to determine whether this series converges, since this is a special case. The root test also does not apply in this scenario.

According the the P-series Test, $\sum_{x=1}^{\infty}\frac{1}{x^p}$ must converge only if . Therefore this could be a valid test, but a wrong definition as the answer choice since the series diverge for $p \leq1$ .

This leaves us with the Integral Test.

$\int_{1}^{\infty}\frac{1}{x}= \lim_{b \to \infty}\int_{1}^{b}\frac{1}{x}dx=\lim_{b \to \infty}( ln(b)-ln(1))=\infty$

Since the improper integral diverges, so does the series.

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Question

Determine if the series converges or diverges. You do not need to find the sum.

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

Tap to reveal answer

Answer

We can compare this to the series $\sum_{n = 1}^{\infty } \frac{1}{n^{3}}$ which we know converges by the p-series test.

To figure this out, let's first compare $n^{3} + 2$ to $n^{3}$ . For any number n, $n^{3} + 2$ will be larger than $n^{3}$ .

There is a rule in math that if you take the reciprocal of each term in an inequality, you are allowed to flip the signs.

Thus, $n^{3} + 2 > n^{3}}$ turns into

$\frac{1}{n^{3}} > \frac{1}{n^{3} + 2}$ .

And so, because $\frac{1}{n ^{3}}$ converges, thus our series also converges.

← Didn't Know|Knew It →

Question

True or False, a -series cannot be tested conclusively using the ratio test.

Tap to reveal answer

Answer

We cannot test for convergence of a -series using the ratio test. Observe,

For the series $\sum_{n=1}^\infty \frac{1}{n^p}$ ,

$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} | \frac{n^p}{(n+1)^p}| = \lim_{n \to \infty} (\frac{n}{n+1})^p = 1^p =1$ .

Since this limit is regardless of the value for , the ratio test is inconclusive.

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Question

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

Tap to reveal answer

Answer

$\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$ 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 , $\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$ hence also converges by the comparison test.

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Question

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

$\frac{8}{\sqrt{n(n+1)(n+2)}}$

Tap to reveal answer

Answer

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

We need to note that the following limit

$\frac{\frac{8}{\sqrt{n(n+1)(n+2)}}}{\frac{8}{n^{3/2}}}}$

goes to 1 as n goes to infinity.

Therefore the series have the same nature. They either converge or diverge at the same time.

We will focus on the series:

$\frac{1}{n^{3/2}}$ .

We know that this series is convergent because it is a p-series. (Remember that

$\sum_{n=1}^{\infty}\frac{1}{n^p}$ converges if p>1 and we have p=3/2 which is greater that one in this case)

By the Limit Comparison Test, we deduce that the series is convergent, and that is what we needed to show.

← Didn't Know|Knew It →

Question

Does the series $\sum_{n=0}^{\infty }\frac{(-1)^n}{n}$ converge conditionally, absolutely, or diverge?

Tap to reveal answer

Answer

The series converges conditionally.

The absolute values of the series $\sum_{n=0}^{\infty }\left |\frac{(1)^n}{n} \right |$ is a divergent p-series with $p\leq 1$ .

However, the the limit of the sequence $\lim_{n\rightarrow \infty }\frac{(-1)^n}{n}=0$ and it is a decreasing sequence.

Therefore, by the alternating series test, the series converges conditionally.

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Question

Which of the following tests will help determine whether $\sum_{x=1}^{\infty}\frac{1}{x}$ is convergent or divergent, and why?

Tap to reveal answer

Answer

The series $\sum_{x=1}^{\infty}\frac{1}{x}$ is a harmonic series.

The Nth term test and the Divergent test may not be used to determine whether this series converges, since this is a special case. The root test also does not apply in this scenario.

According the the P-series Test, $\sum_{x=1}^{\infty}\frac{1}{x^p}$ must converge only if . Therefore this could be a valid test, but a wrong definition as the answer choice since the series diverge for $p \leq1$ .

This leaves us with the Integral Test.

$\int_{1}^{\infty}\frac{1}{x}= \lim_{b \to \infty}\int_{1}^{b}\frac{1}{x}dx=\lim_{b \to \infty}( ln(b)-ln(1))=\infty$

Since the improper integral diverges, so does the series.

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Question

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

Tap to reveal answer

Answer

$\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$ 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 , $\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$ hence also converges by the comparison test.

← Didn't Know|Knew It →

Question

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

$\frac{8}{\sqrt{n(n+1)(n+2)}}$

Tap to reveal answer

Answer

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

We need to note that the following limit

$\frac{\frac{8}{\sqrt{n(n+1)(n+2)}}}{\frac{8}{n^{3/2}}}}$

goes to 1 as n goes to infinity.

Therefore the series have the same nature. They either converge or diverge at the same time.

We will focus on the series:

$\frac{1}{n^{3/2}}$ .

We know that this series is convergent because it is a p-series. (Remember that

$\sum_{n=1}^{\infty}\frac{1}{n^p}$ converges if p>1 and we have p=3/2 which is greater that one in this case)

By the Limit Comparison Test, we deduce that the series is convergent, and that is what we needed to show.

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Question

Does the series $\sum_{n=0}^{\infty }\frac{(-1)^n}{n}$ converge conditionally, absolutely, or diverge?

Tap to reveal answer

Answer

The series converges conditionally.

The absolute values of the series $\sum_{n=0}^{\infty }\left |\frac{(1)^n}{n} \right |$ is a divergent p-series with $p\leq 1$ .

However, the the limit of the sequence $\lim_{n\rightarrow \infty }\frac{(-1)^n}{n}=0$ and it is a decreasing sequence.

Therefore, by the alternating series test, the series converges conditionally.

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Question

Which of the following tests will help determine whether $\sum_{x=1}^{\infty}\frac{1}{x}$ is convergent or divergent, and why?

Tap to reveal answer

Answer

The series $\sum_{x=1}^{\infty}\frac{1}{x}$ is a harmonic series.

The Nth term test and the Divergent test may not be used to determine whether this series converges, since this is a special case. The root test also does not apply in this scenario.

According the the P-series Test, $\sum_{x=1}^{\infty}\frac{1}{x^p}$ must converge only if . Therefore this could be a valid test, but a wrong definition as the answer choice since the series diverge for $p \leq1$ .

This leaves us with the Integral Test.

$\int_{1}^{\infty}\frac{1}{x}= \lim_{b \to \infty}\int_{1}^{b}\frac{1}{x}dx=\lim_{b \to \infty}( ln(b)-ln(1))=\infty$

Since the improper integral diverges, so does the series.

← Didn't Know|Knew It →

Question

Determine if the series converges or diverges. You do not need to find the sum.

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

Tap to reveal answer

Answer

We can compare this to the series $\sum_{n = 1}^{\infty } \frac{1}{n^{3}}$ which we know converges by the p-series test.

To figure this out, let's first compare $n^{3} + 2$ to $n^{3}$ . For any number n, $n^{3} + 2$ will be larger than $n^{3}$ .

There is a rule in math that if you take the reciprocal of each term in an inequality, you are allowed to flip the signs.

Thus, $n^{3} + 2 > n^{3}}$ turns into

$\frac{1}{n^{3}} > \frac{1}{n^{3} + 2}$ .

And so, because $\frac{1}{n ^{3}}$ converges, thus our series also converges.

← Didn't Know|Knew It →

Question

True or False, a -series cannot be tested conclusively using the ratio test.

Tap to reveal answer

Answer

We cannot test for convergence of a -series using the ratio test. Observe,

For the series $\sum_{n=1}^\infty \frac{1}{n^p}$ ,

$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} | \frac{n^p}{(n+1)^p}| = \lim_{n \to \infty} (\frac{n}{n+1})^p = 1^p =1$ .

Since this limit is regardless of the value for , the ratio test is inconclusive.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

$\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$ 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 , $\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$ hence also converges by the comparison test.

← Didn't Know|Knew It →

Question

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

$\frac{8}{\sqrt{n(n+1)(n+2)}}$

Tap to reveal answer

Answer

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

We need to note that the following limit

$\frac{\frac{8}{\sqrt{n(n+1)(n+2)}}}{\frac{8}{n^{3/2}}}}$

goes to 1 as n goes to infinity.

Therefore the series have the same nature. They either converge or diverge at the same time.

We will focus on the series:

$\frac{1}{n^{3/2}}$ .

We know that this series is convergent because it is a p-series. (Remember that

$\sum_{n=1}^{\infty}\frac{1}{n^p}$ converges if p>1 and we have p=3/2 which is greater that one in this case)

By the Limit Comparison Test, we deduce that the series is convergent, and that is what we needed to show.

← Didn't Know|Knew It →

Knew It

P-Series - AP Calculus BC

Determine if the series converges or diverges. You do not need to find the sum.

True or False, a -series cannot be tested conclusively using the ratio test.

We cannot test for convergence of a -series using the ratio test. Observe, For the series , . Since this limit is regardless of the value for , the ratio test is inconclusive.

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

converges due to the comparison test. We start with the equation . Since for all values of k, we can multiply both side of the equation by the inequality and get for all values of k. Since is a convergent p-series with , hence also converges by the comparison test.

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

Does the series converge conditionally, absolutely, or diverge?

The series converges conditionally. The absolute values of the series is a divergent p-series with . However, the the limit of the sequence and it is a decreasing sequence. Therefore, by the alternating series test, the series converges conditionally.

Which of the following tests will help determine whether is convergent or divergent, and why?

Determine if the series converges or diverges. You do not need to find the sum.

True or False, a -series cannot be tested conclusively using the ratio test.

We cannot test for convergence of a -series using the ratio test. Observe, For the series , . Since this limit is regardless of the value for , the ratio test is inconclusive.

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

converges due to the comparison test. We start with the equation . Since for all values of k, we can multiply both side of the equation by the inequality and get for all values of k. Since is a convergent p-series with , hence also converges by the comparison test.

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

Does the series converge conditionally, absolutely, or diverge?

The series converges conditionally. The absolute values of the series is a divergent p-series with . However, the the limit of the sequence and it is a decreasing sequence. Therefore, by the alternating series test, the series converges conditionally.

Which of the following tests will help determine whether is convergent or divergent, and why?

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

converges due to the comparison test. We start with the equation . Since for all values of k, we can multiply both side of the equation by the inequality and get for all values of k. Since is a convergent p-series with , hence also converges by the comparison test.

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

Does the series converge conditionally, absolutely, or diverge?

The series converges conditionally. The absolute values of the series is a divergent p-series with . However, the the limit of the sequence and it is a decreasing sequence. Therefore, by the alternating series test, the series converges conditionally.

Which of the following tests will help determine whether is convergent or divergent, and why?

Determine if the series converges or diverges. You do not need to find the sum.

True or False, a -series cannot be tested conclusively using the ratio test.

We cannot test for convergence of a -series using the ratio test. Observe, For the series , . Since this limit is regardless of the value for , the ratio test is inconclusive.

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

converges due to the comparison test. We start with the equation . Since for all values of k, we can multiply both side of the equation by the inequality and get for all values of k. Since is a convergent p-series with , hence also converges by the comparison test.

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

Determine if the series converges or diverges. You do not need to find the sum.

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

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

$\frac{8}{\sqrt{n(n+1)(n+2)}}$

Does the series $\sum_{n=0}^{\infty }\frac{(-1)^n}{n}$ converge conditionally, absolutely, or diverge?

Which of the following tests will help determine whether $\sum_{x=1}^{\infty}\frac{1}{x}$ is convergent or divergent, and why?

Determine if the series converges or diverges. You do not need to find the sum.

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

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

$\frac{8}{\sqrt{n(n+1)(n+2)}}$

Does the series $\sum_{n=0}^{\infty }\frac{(-1)^n}{n}$ converge conditionally, absolutely, or diverge?

Which of the following tests will help determine whether $\sum_{x=1}^{\infty}\frac{1}{x}$ is convergent or divergent, and why?

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

$\frac{8}{\sqrt{n(n+1)(n+2)}}$

Does the series $\sum_{n=0}^{\infty }\frac{(-1)^n}{n}$ converge conditionally, absolutely, or diverge?

Which of the following tests will help determine whether $\sum_{x=1}^{\infty}\frac{1}{x}$ is convergent or divergent, and why?

Determine if the series converges or diverges. You do not need to find the sum.

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

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

$\frac{8}{\sqrt{n(n+1)(n+2)}}$