Change a variable to transform a series into a power series

[whitejac]

Homework Statement

The following series are not power series, but you can transform each one into a power
series by a change of variable and so find out where it converges.

∑^∞₀ ((3ⁿ(n+1)) / (x+1)ⁿ

Homework Equations

a power series is a series of the form:

a0 + a1x + a2x^2 ... + ...

The Attempt at a Solution

What exactly does it mean by transforming a power series by changing a variable? the only thing I could look up and find was converting to another coordinate plane in calculus 3 which would be further along than this problem focuses on... but i am still confused as i can't see how a series can be in a separate plane like that

 

[S.G. Janssens]

Introduce a new variable $y$ as a function of $x$, such that if you write your series in terms of $y$ it takes the form of a power series. Then determine for which values of $y$ that power series converges. Finally, translate these values back to values of $x$.

 

[whitejac]

Let me see if i understand this then...

Let Y = Y(x),
Where Y(x) = 1 / (x - 1)

Then we would have:

3ⁿ(n+1) / ( 1/ (x -1 +1)ⁿ)
= 3ⁿ(n+1) (x)ⁿ

this would in turn be a power series, yes?

 

[Mark44]

Let me see if i understand this then...

Let Y = Y(x),
Where Y(x) = 1 / (x - 1)

No, this isn't the substitution. There is another that should be much more obvious.

Then we would have:

3ⁿ(n+1) / ( 1/ (x -1 +1)ⁿ)
= 3ⁿ(n+1) (x)ⁿ

this would in turn be a power series, yes?

You have a series whose general term is $3^n(n + 1) \cdot \frac 1 {(x + 1)^n}$. You'd like to end up with $3^n(n + 1) \cdot y^n$. What's the most obvious substitution to make?

 

[whitejac]

how is mine not obvious enough? It removes the added 1 and puts y back in the numerator.

 

[whitejac]

Sorry, I meant for my original definition of y to be:
x = 1/y -1
then you would have the relationship I intended to have:

3n(n+1)⋅yn

 

[Mark44]

Sorry, I meant for my original definition of y to be:
x = 1/y -1

That works, but why not write it as $y = \frac 1 {x + 1}$?
Then $y^n = \frac 1 {(x + 1)^n}$

 

[whitejac]

So, essentially, my answer could work but you could also simply say "Let y = all of the business that makes this difficult"?

 

[Mark44]

So, essentially, my answer could work but you could also simply say "Let y = all of the business that makes this difficult"?

No, not really.

Sorry, I meant for my original definition of y to be:
x = 1/y -1

When you define something (in this case, y), you generally start it with "Let y = <whatever> ". What you wrote is the inverse of the function that defines y.

 

[whitejac]

No, not really.

When you define something (in this case, y), you generally start it with "Let y = <whatever> ". What you wrote is the inverse of the function that defines y.

Yeah, I noticed that. I guess it makes sense. Thank you!