Interval of Convergence for Power Series

[ineedhelpnow]

Hi hi,
So I worked on this problem and I know I probably made a mistake somewhere towards the end so I was hoping one of you would catch it for me. Thank you!

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Pasteboard — Uploaded Image

 

[MarkFL]

According to W|A, the series indeed converges for:

$$|-x-1|<3$$

which gives:

$$-4<x<2$$

just as you found. (Yes)

By the way, I am going to move this thread to our Calculus forum as its a better fit I think. (Thinking)

 

[ineedhelpnow]

Thank you ^^ I wasn't sure if it would be -3<-x-1<3 or -3<x+1<3

 

[MarkFL]

Thank you ^^ I wasn't sure if it would be -3<-x-1<3 or -3<x+1<3

Note that:

$$|-x-1|<3$$

is the same as:

$$|x+1|<3$$

which can be written as:

$$\sqrt{(x-(-1))^2}<3$$

and this tells us (because it is a 1-dimensional distance formula) that we want all real $x$ whose distance (on the number line) from $-1$ is less than $3$, and so we have:

$$-1-3<x<-1+3$$

or

$$-4<x<2$$

You should also observe that either inequality you posted will lead to the correct answer:

i) $$-3<-x-1<3$$

Add though by 1:

$$-2<-x<4$$

Multiply through by -1:

$$2>x>-4$$

And this is equivalent to:

$$-4<x<2$$

ii) $$-3<x+1<3$$

Subtract through by 1:

$$-4<x<2$$

 

[ineedhelpnow]

I noticed they both come to the same solution but I wasn't sure if there was a "right" way to have it done. Thanks a ton for the thorough explanation!