This is an alternating series test.

In order to find the terms necessary to approximate the series within  first see if the series is convergent using the alternating series test. If the series converges, find n such that 

Step 1:

An alternating series can be identified because terms in the series will “alternate” between + and –, because of 

Note: Alternating Series Test can only show convergence. It cannot show divergence.

If the following 2 tests are true, the alternating series converges.

1.        
2.        {} is a decreasing sequence, or in other words 

Solution:

1.

2.

Since the 2 tests pass, this series is convergent.

Step 2:

Plug in n values until 

7 terms are needed to approximate the sum within .001

This is an alternating series test.

In order to find the terms necessary to approximate the series within  first see if the series is convergent using the alternating series test. If the series converges, find n such that 

Step 1:

An alternating series can be identified because terms in the series will “alternate” between + and –, because of 

Note: Alternating Series Test can only show convergence. It cannot show divergence.

If the following 2 tests are true, the alternating series converges.

1.       
2.        {} is a decreasing sequence, or in other words 

Solution:

1.

2. {{b_{n}} is a decreasing sequence. A factorial always increases as n increases, so each term will decrease as n increases.

Since the 2 tests pass, this series is convergent.

Step 2:

Plug in n values until 

4 terms need to be added to approximate the sum within .001

To determine whether the given alternating series converges or diverges, we must perform the Alternating Series test, which states that for a given series

 and  or , for the series to converge,  and  must be decreasing.

To start, we must take the limit of  as n approaches infinity:

because (the numerator goes to zero).

Next, we must see if  is decreasing. Simply increasing  to  does not clearly show whether the function is decreasing because both the numerator and denominator increase. So, we must find the first derivative and see if it is negative:

,

and was found using the following rules:

,

The derivative is always negative from  to , so the sequence   is decreasing. The series is (absolutely) convergent because it passed both parts of the test.

To determine whether the series is convergent or divergent, we must use the Alternating Series test, which states that for a given series , where  or , if  and  is decreasing, then the series is convergent.

First, we must evaluate the limit of  as  approaches infinity:

Therefore, the test fails and series is divergent.

To determine whether the series is convergent or divergent, we must use the Alternating Series test, which states that for a given series , where  or , if  and  is decreasing, then the series is convergent.

First, we must evaluate the limit of  as  approaches infinity:

The test fails and the series is therefore divergent. 

To determine whether the series is convergent or divergent, we must use the Alternating Series test, which states that for a given series , where  or , if  and  is decreasing, then the series is convergent.

First, we must evaluate the limit of  as  approaches infinity:

The test fails and therefore the series is divergent. 

Recall the Maclaurin series formula:

Despite being a 5th degree polynomial recall that the Maclaurin series for any polynomial is just the polynomial itself, so this function's Taylor series is identical to itself with two non-zero terms.

The only function that has four or fewer terms is  as its Maclaurin series is.

Using the formula of a binomial series centered at 0:

 ,

where we replace  with  and , we get:

  for the first 3 terms.

Then, we find the terms where,

By the ratio test, the series  converges absolutely: 

Using the root test

and

. T

herefore, the series only converges when it is equal to zero.

This occurs when x=5.