Help with calculus problems 2 and 4

[Shay10825]

Hello. I need help on these 4 problems. I wrote the problems out and showed my work for numbers 1 and 3. I have no clue how to do numbers 2 and 4. Any help would be greatly appreciated.

[img=http://img10.imageshack.us/img10/8426/calcth31tq.th.jpg]

Thanks

 

[HallsofIvy]

Your solution to number 1 looks good.

For number 2: DO it! Let x= 1 in the power series. What do you get?
What is ln(1)? (The $\int \frac{1}{x}dx$ is an indefinite integral- it makes no sense to "let x= 1" there so just ignore it.)

For number 3, did you check the endpoints of the interval?

For number 4, the sum in 2 is alternating, + and -. Each partial sum lies BETWEEN the two previous ones so the infinite sum lies BETWEEN any two consecutive partial sums. Since you are asked to use only the first four terms, Calculate the sum of the first three terms, then the sum of the first four,with x= 1.2. The true value must lie between those values.

 

[Shay10825]

For number 2: DO it! Let x= 1 in the power series. What do you get? What is ln(1)?

I get 0 but how does that prove that C=0?

 

[Shay10825]

For number 3, did you check the endpoints of the interval?

When x= 0 i get:
-1 - 1/2 - 1/3 - 1/4 . . .
so it converges at 0

When x=2 i get:
1- 1/2 + 1/3 - 1/4 . . .
so it converges at 2 so then the answer would be [0,2] right?

 

[HallsofIvy]

I get 0 but how does that prove that C=0?

You get 0 for two different things! ln(1)= 0 and the series on the right is just a sum of 0 for x= 1. What does the equation 0= C+ 0 tell you?

 

[HallsofIvy]

When x= 0 i get:
-1 - 1/2 - 1/3 - 1/4 . . .
so it converges at 0

No! It doesn't!

When x=2 i get:
1- 1/2 + 1/3 - 1/4 . . .
so it converges at 2 so then the answer would be [0,2] right?

 

[Shay10825]

No! It doesn't!

How do you know that it diverges?

 

[Shay10825]

is what I did for number 4 correct?

[img=http://img123.imageshack.us/img123/3203/calcth56wo.th.jpg]