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美联社Calculus BC : P-Series

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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:

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

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

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

None of the others converge.

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

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 #1 :P Series

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

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

Possible Answers:

The series is convergent.

$\frac{\pi}{5}$

$\frac{\pi}{7}$

The series is divergent.

$\frac{1}{2}$

Correct answer:

The series is convergent.

Explanation:

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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Example Question #1 :P Series

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

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

Possible Answers:

Diverges

Conditionally converges.

无论是收敛还是发散的。

There is not enough information to decide convergence.

Converges

Correct answer:

Converges

Explanation:

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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Example Question #1 :Harmonic Series

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

Possible Answers:

Integral Test: The improper integral determines that the harmonic series diverge.

$\textup{}$ Nth Term Test: The series diverge because the limit asgoes to infinity is zero.

P-Series Test: The summation converges since p=1 .

$\textup{}$ Divergence Test: Since limit of the series approaches zero, the series must converge.

Root Test: Since the limit asapproaches to infinity is zero, the series is convergent.

Correct answer:

Integral Test: The improper integral determines that the harmonic series diverge.

Explanation:

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 p>1 . 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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Example Question #1 :P Series

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

Possible Answers:

Converge Absolutely.

Does not exist.

Converge Conditionally.

Diverges.

Cannot tell with the given information.

Correct answer:

Converge Conditionally.

Explanation:

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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Example Question #1 :P Series

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

Possible Answers:

False

True

Correct answer:

True

Explanation:

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 isregardless of the value for, the ratio test is inconclusive.

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美联社Calculus BC : P-Series

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #1 :Concepts Of Convergence And Divergence

Example Question #1 :P Series

Example Question #1 :P Series

Example Question #1 :Harmonic Series

Example Question #1 :P Series

Example Question #1 :P Series

All AP Calculus BC Resources

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