To determine the convergence of the series, we must use the ratio test, which states that for the series , and , if L is greater than 1, the series diverges, if L is equal to 1, the series may absolutely converge, conditionally converge, or diverge, and if L is less than 1 the series is (absolutely) convergent.

For our series, we get

Using the properties of radicals and exponents to simplify, we get

L is greater than 1 so the series is divergent.

To determine the convergence of the series, we must use the ratio test, which states that for the series , and , if L is greater than 1, the series diverges, if L is equal to 1, the series may absolutely converge, conditionally converge, or diverge, and if L is less than 1 the series is (absolutely) convergent.

For our series, we get

Using the properties of radicals and exponents to simplify, we get

L is less than 1 so the series is convergent.

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The ratio test can be used to prove that an infinite series  is convergent or divergent. In some cases, however, the ratio test may be inconclusive. 

Define:

 If...

 then the series is absolutely convergent, and therefore convergent. 

 then the series is divergent, 

 then the series is either conditionally convergent, absolutely convergent, or divergent.  

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To compute the limit, we first need to write the expression for . 

Now we can find the limit, 

We can cancel all factors with the  exponent. 

Now the negative factor in the numerator   will cancel with  in the denominator, although this is trivial since we are taking the absolute value regardless. 

Continue simplifying, 

Therefore, 

This proves that the series  is absolutely convergent, and therefore convergent. 

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The ratio test can be used to prove that an infinite series  is convergent or divergent. In some cases, however, the ratio test may be inconclusive. 

Define:

 If...

 then the series is absolutely convergent, and therefore convergent. 

 then the series is divergent, 

 then the series is either conditionally convergent, absolutely convergent, or divergent. 

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Divide above and below by , in the limit the all terms disappear except for the coefficients on the leading terms. The limit is therefore equal to 

Therefore, the series is either conditionally convergent, absolutely convergent, or divergent. 

We will use the integral test to prove this result.

We need to note the following:

    is positive, decreasing and .

By the integral test, we know that the series  is the integral .

We know that the above intgral is finite.

This means that the series

is convergent.

We know that:

and therefore we deduce :

We will use the Comparison Test with this problem. To do this we will look at the function in general form . 

We can do this since,

 and   approach zero as n approaches infinity. The limit of our function becomes,

This last part gives us .

Now we know that and noting that is a geometric series that is convergent.

We deduce by the Comparison Test that the series

having general term is convergent.

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 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.

To have a series that is convergent we must have that the general term of the series goes to 0 as n goes to .

We have the general term:

therefore, we have .

This means that the general term does not go to 0 .

Therefore the series is divergent.