We note first that the general term of the series is positive.

It is also decreasing and tends to 0 as n tends to . We will use the Integral Comparison Test to show this result.

Note the nature of the series is the same for the integral:

This last intgral is divergent because it does not equal zero.

Therefore our series is divergent as well.

We note first that we can write the general term as: 

and simplifying this term one more time, we have:

We note that since ,  this series is a geometric one which is convergent.

This is what we need to show here.

We know that if a series is convergent, then its general term must go to 0 as  .

We have  is our general term in this case.

We have .

Since the general term does not go to 0, the series is divergent.

We will use the Limit Comparison Test to study the nature of the series.

We note first that , the series is positive.

We will compare the general term to . 

We note that by letting  and , we have:

.

Therefore the two series have the same nature, (they either converge or diverge at the same time). 

We will use the Integral Test to deduce that the series having the general term:

is convergent.

Note that we know that is convergent if p>1 and in our case p=8 .

This shows that the series having general term is convergent.

By the Limit Test, the series having general term  is convergent.

This shows that our series is convergent.

We will use the Comparison Test to prove this result.

We need to note first that  for .

We know that  , where .

Inverting the above inequality, we have:

.

Now we will use the Comparison Test.

We know that the series is divergent.

Therefore,

is also divergent.

We will use the Comparison Test to prove this result. We must note the following:

  is positive. 

We have all natural numbers n:

, this implies that

.

Inverting we get :

Summing from 1 to , we have

We know that the is divergent. Therefore by the Comparison Test:

 is divergent.

We will use the comparison test to prove that

converges (Note: we cannot use the ratio test, because then the ratio will be , which means the test is inconclusive).

We will compare  to  because they "behave" somewhat similarly. Both series are nonzero for all , so one of the conditions is satisfied.

The series 

converges, so we must show that 

for .

This is easy to show because

since the denominator  is greater than or equal to  for all .

Thus, since 

and because

converges, it follows that 

converges, by comparison test.

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 can solve this problem quite simply with the integral test. We know that if 

converges, then our series converges. 

We can rewrite the integral as 

and then use our formula for the antiderivative of power functions to get that the integral equals

.

We know that this only goes to zero if . Subtracting p from both sides, we get

.

We compare this series to the series 

Because 

  for   

it follows that

  for   

This implies

Because the series on the right has an exponent ,

the series on the right converges

making

converge as well.