Finding interval of convergence

[physics=world]

1. Determine a power series, centered at zero for the function ∫f(x)dx. Identify the interval of convergence.

f(x) = ln(x+1) = ∫$\frac{1}{x+1}$

2.


3. i found the power series, which is :

Ʃ ((-1)^(n))(x^(n+1)) / (n+1)


Im okay with that, but i need help on finding the interval of convergence...


to find it i do ratio test.. and i get


-1< x < 1


I need help on checking for the endpoints

the answer in the book is (-1,1]


how do i know that 1 converges? do i just plug in 1 into the function?

 

[Dick]

1. Determine a power series, centered at zero for the function ∫f(x)dx. Identify the interval of convergence.

f(x) = ln(x+1) = ∫$\frac{1}{x+1}$

2.


3. i found the power series, which is :

Ʃ ((-1)^(n))(x^(n+1)) / (n+1)


Im okay with that, but i need help on finding the interval of convergence...


to find it i do ratio test.. and i get


-1< x < 1


I need help on checking for the endpoints

the answer in the book is (-1,1]


how do i know that 1 converges? do i just plug in 1 into the function?

No, plug 1 into the series. See if the series converges. Do the same thing at x=(-1).

 

[physics=world]

okay so when i plug in 1 into the power series that i found..

i get

Ʃ (-1^n)(1^(n+1)) / (n+1)

how do i know if this converges or diverges?

do i use one of the series test?

 

[Dick]

okay so when i plug in 1 into the power series that i found..

i get

Ʃ$\frac{(-1^n)(1^(n+1)}{n+1}$

how do i know if this converges or diverges?

do i use one of the series test?

Sure, use a series test. 1^(n+1)=1. Looks like the alternating series test might be a good one.

 

[physics=world]

hmm when i use the alternating series test is my a_n = (1^(n+1)) / (n+1)

because when i use that for a_n

i get (1^(infinity)) / (infinity)

or do we ignore (1^(n+1)) because it will always equal to 1?

 

[Dick]

hmm when i use the alternating series test


is my a_n = (1^(n+1)) / (n+1)

because when i use that for a_n

i get (1^(infinity)) / (infinity)

That is NOT the alternating series test. State the alternating series test.

 

[physics=world]

isnt it the lim as n -> infinity of a_n = 0

and

0 < a_n+1 ≤ a_n

 

[Dick]

isnt it the lim as n -> infinity of a_n = 0

and

0 < a_n+1 ≤ a_n

That's part of the premises. State the rest including the conclusion.

 

[physics=world]

That's part of the premises. State the rest including the conclusion.

conclusion?

 

[Dick]

conclusion?

The conclusion is that if a series satisfies certain conditions then it converges. Does your series satisfy all of the conditions.

 

[physics=world]

yes the series does satisfy the conditions for 1.

 

[Dick]

yes the series does satisfy the conditions for 1.

Alright. So it converges at x=1. Now try x=(-1).

 

[physics=world]

the lim as n -> infinity of (-1^(n+1)) / (n+1)

im i using the alternate series test correct for this one?

 

[Dick]

the lim as n -> infinity of (-1^(n+1)) / (n+1)

im i using the alternate series test correct for this one?

No, you left out the (-1)^n factor. Write out the first few terms of the series. Does it really alternate?

 

[physics=world]

okay. so it does not alternate it just stays negative

 

[Dick]

okay. so it does not alternate it just stays negative

Ok, then you need another test. Does it look like it converges to you?

 

[physics=world]

Ok, then you need another test. Does it look like it converges to you?

No it does not look like it will converge

 

[Dick]

No it does not look like it will converge

If you can give a good reason why then then you are done. It looks pretty similar to the harmonic series 1/n to me.