P-Series

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AP Calculus BC › P-Series

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

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

CORRECT

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

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

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

None of the others converge.

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

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

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

Converges

CORRECT

Diverges

There is not enough information to decide convergence.

Conditionally converges.

Neither converges nor diverges.

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.

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

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

The series is convergent.

CORRECT

The series is divergent.

$\frac{1}{2}$

$\frac{\pi}{7}$

$\frac{\pi}{5}$

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.

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

Converge Conditionally.

CORRECT

Converge Absolutely.

Diverges.

Cannot tell with the given information.

Does not exist.

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.

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

True

CORRECT

False

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

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

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

CORRECT

P-Series Test: The summation converges since .

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

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

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

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 . Therefore this could be a valid test, but a wrong definition as the answer choice since the series diverge for $p \leq1$ .