Given just the harmonic series, we would state that the series diverges. However, we are given the alternating harmonic series. To determine whether this series will converge or diverge, we must use the Alternating Series test. 

The test states that for a given series where  or  where  for all n, if  and  is a decreasing sequence, then  is convergent.

First, we must evaluate the limit of  as n approaches infinity:

The limit equals zero because the numerator of the fraction equals zero as n approaches infinity. 

Next, we must determine if  is a decreasing sequence. , thus the sequence is decreasing.

Because both parts of the test passed, the series is (absolutely) convergent.

We can use the alternating series test to show that

converges.

We must have   for  in order to use this test. This is easy to see because  is in for all  (the values of this sequence are ), and sine is always nonzero whenever sine's argument is in .

Now we must show that

1. 

2.  is a decreasing sequence.

The limit 

implies that 

so the first condition is satisfied.

We can show that  is decreasing by taking its derivative and showing that it is less than  for :

The derivative is less than , because  is always less than , and that  is positive for , using a similar argument we used to prove that  for . Since the derivative is less than ,  is a decreasing sequence. Now we have shown that the two conditions are satisfied, so we have proven that 

converges, by the alternating series test.

The series given is an alternating series.  

Write the three rules that are used to satisfy convergence in an alternating series test.

For :

The first and second conditions are satisfied since the terms are positive and are decreasing after each term.

However, the third condition is not valid since  and instead approaches infinity.

The correct answer is:

To determine whether the series converges or diverges, we must use the Alternating Series test, which states that for 

 - and  where  for all n - to converge, 

 must equal zero and  must be a decreasing series.

For our series, 

because it behaves like 

.

The test fails because  so we do not need to check the second condition of the test.

The series is divergent.

We can show that the series   diverges using the ratio test.

 will dominate over  since it's a higher order term. Clearly, L will not be less than, which is necessary for absolute convergence. 

Alternatively, it's clear that  is much greater than , and thus having  in the numerator will make the series diverge by the  limit test (since the terms clearly don't converge to zero).

The other series will converge by alternating series test, ratio test, geometric series, and comparison tests.

 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.

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. 

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.

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.