Explanation:

We compare this series to the series

Because

for

it follows that

for

This implies

Because the series on the right has a degree ofequal toin the denominator,

the series on the right diverges

making

diverge as well.

Explanation:

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 multiplyingby, we obtain.

Since this holds for allwe can conclude that

.

Summing fromto, 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

.

Explanation:

Notice thatfor

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.

Explanation:

Explanation:

Explanation:

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

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

convergesas 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.

Explanation:

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

Explanation:

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:

具体地说:

This limit equals one because of the fact that:

if the coefficients come from the same power.

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

Explanation:

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

Therefore, said sum converges.

Then, by comparison test,also converges.

Explanation:

Sincefor all n>0, andconverges as a P-Series, we may conclude thatmust also converge by the comparison test.