We compare this series to the series 

Because 

  for   

it follows that

  for   

This implies

Because the series on the right has a degree of  equal to  in the denominator,

the series on the right diverges

making

diverge as well.

To show this series converges, we use direct comparison with

,

which converges by the p-series test with .

Thus we must show that

.

Cross multiplying the previous section and multiplying by , we obtain .

Since this holds for all we can conclude that 

.

Summing from  to , and noting that 

for all , we obtain the following inequality:

.

Therefore the series 

converges by direct comparison.

Now to find the value, we note that 

,

so that

.

Now let 

be a sequence of partial sums.

Then we have 

Therefore

.

Taking the limit as , we obtain the following:

Therefore we have

.

Notice that for

This implies that

for

Which then  implies

Since the right-hand side is the harmonic series, we have

and thus the series does NOT converge.

Step 1: Recall the convergence rule of the power series:

According to the convergence rule of the power series....

 converges as long as 

 Step 2: Compare the exponent:

Since , it is greater than . Hence the series converges.

Step 3: Conclusion of the convergence rule

Now, notice that the series isn't an alternating series, so it doesn't matter whether we check for absolute or conditional convergence.

Step 1: Try and look for another function that is similar to the original function:

 looks like 

Step 2: We will now Use the Limit Comparison test

Since the limit calculated, is not equal to 0, the given series converges by limit comparison test

The best way to answer this question would be by comparing the series to another series,, that greatly resembles the behavior of the original series, . The behavior is determined by the terms of the numerator and the denominator that approach infinity at the quickest rate. In this case:

When this series is simplifies, it simplifies to a series that converges because of the p-test where .

With two series and the confirmed convergence of one of those series, the limit comparison test can be applied to test for the convergence or divergence of the original series. The limit comparison test states that two series will converge or diverge together if:

Specifically:

This limit equals one because of the fact that:

  if the coefficients come from the same power. 

Because the limit is larger than zero,  and  will converge or diverge together. Since it was already established that  converges, the original seies, , converges by the limit comparison test.

 for all ;     is a sum of geometric sequence with base 1/3.

Therefore, said sum converges.

Then, by comparison test,    also converges.

Since  for all n>0, and  converges as a P-Series, we may conclude that  must also converge by the comparison test.