To determine the convergence or divergence of the following series, we must use the Ratio Test, which states that for the series , and , that when , the series diverges, when , the series converges, and when , the series may be absolutely convergent, conditionally convergent, or divergent.

For our series, using the above formula, we get 

which simplified becomes

Note that the factorial was simplified by rewriting .

In the limit, the n in the numerator goes to infinity, which makes L go to infinity, which is greater than 1. Thus, the series diverges.

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

Now, we must find L:

We simplified the limit using the properties of radicals and exponents. The denominator of the limit becomes infinitely large, so the term goes to zero. L is therefore less than 1, so the series converges. 

Using the ratio test , we get . Evaluating the limit, we get . Because  the series converges. Also, because no n value could make the series negative, it absolutely converges . 

Using the ratio test, we get  and therefore, the series diverges. We simplify the inside by saying  and that  which cancels out with the denominator. 

To determine the convergence or divergence of the series, the most apparent way to do so is by using the ratio test. The ratio test states that a series will converge if:

The series will diverge if:

The ratio test can be used by first writing all n as n+1 in the numerator and the denominator is the normal series:

This simplifies to:

Because the limit is larger than one, the series diverges. 

To determine the convergence or divergence of the series, the most apparent way to do so is by using the ratio test. The ratio test states that a series will converge if:

The series will diverge if:

The ratio test can be used by first writing all n as n+1 in the numerator and the denominator is the normal series:

This simplifies to:

Because the limit is larger than one, the series diverges. 

Consider the following limit:

Therefore, 

Then, according to limit ratio test for positive series:

K=1, therefore both  converge or diverge.

We know, that  is generalized harmonic series with p=3, therefore it converges. Consecutively, by limit ratio test,  also converges.

Compute the limit  to determine if the series

  converges or diverges, or neither. 

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Define for the sequence  the limit  . The ratio test states:

the series  

If  then the series  converges absolutely, and therefore converges. 

If  then the series  diverges. 

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

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To write the numerator in the limit we need to compute, note that every  will become  which simplifies down to . So we have for , 

Because  we can conclude that the series  converges by the ratio test.