P-Series - AP Calculus BC

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.

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

We need to note that the following limit

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:

.

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

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.

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.

The series 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, must converge only if . Therefore this could be a valid test, but a wrong definition as the answer choice since the series diverge for .

This leaves us with the Integral Test.

Since the improper integral diverges, so does the series.

We can compare this to the series which we know converges by the p-series test.

To figure this out, let's first compare to . For any number n, will be larger than .

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, turns into

.

And so, because converges, thus our series also converges.

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.

We can compare this to the series which we know converges by the p-series test.

To figure this out, let's first compare to . For any number n, will be larger than .

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, turns into

.

And so, because converges, thus our series also converges.

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.

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.

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

We need to note that the following limit

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:

.

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

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.

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.

The series 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, must converge only if . Therefore this could be a valid test, but a wrong definition as the answer choice since the series diverge for .

This leaves us with the Integral Test.

Since the improper integral diverges, so does the series.

We can compare this to the series which we know converges by the p-series test.

To figure this out, let's first compare to . For any number n, will be larger than .

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, turns into

.

And so, because converges, thus our series also converges.

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.

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.

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

We need to note that the following limit

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:

.

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

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.

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.

The series 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, must converge only if . Therefore this could be a valid test, but a wrong definition as the answer choice since the series diverge for .

This leaves us with the Integral Test.

Since the improper integral diverges, so does the series.

We can compare this to the series which we know converges by the p-series test.

To figure this out, let's first compare to . For any number n, will be larger than .

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, turns into

.

And so, because converges, thus our series also converges.

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.