Determining General Values of Convergence for a seqence

[trap101]

Determine the values of "r" for which rⁿ converges.


Is there a specific procedure I should try to apply to figure this out? The only things I could intuitively come up with that will converge in this scenario are when -1 ≤ r ≤ 1...is there anything else to this?

 

[LCKurtz]

Really? r = -1?? For |r|<1 I guess it depends on what you have to work with. You might consider what happens to $y=\ln(|r|^n)$.

 

[trap101]

Really? r = -1?? For |r|<1 I guess it depends on what you have to work with. You might consider what happens to $y=\ln(|r|^n)$.

I don't fully see what your getting at. From what you wrote that would become n ln(r), but in that case it would not converge.

 

[LCKurtz]

Really? r = -1?? For |r|<1 I guess it depends on what you have to work with. You might consider what happens to $y=\ln(|r|^n)$.

I don't fully see what your getting at. From what you wrote that would become n ln(r), but in that case it would not converge.

That's "you're", not "your".

What about $r=-1$? Have you thought about that?

Yes, $y=\ln(|r|^n) = n \ln(|r|)$ doesn't converge. But what does it do? Since $y$ is the logarithm of your problem, what does that tell you about your problem?

 

[trap101]

That's "you're", not "your".

What about $r=-1$? Have you thought about that?

Yes, $y=\ln(|r|^n) = n \ln(|r|)$ doesn't converge. But what does it do? Since $y$ is the logarithm of your problem, what does that tell you about your problem?

In the case of r = -1, the sequence oscillates so it is divergent.

In the second part $y=\ln(|r|^n) = n \ln(|r|)$ goes to infinity whenever r > 1, when 0 < r < 1 then it converges.

So the main behavior to try and examine occurs between -1 < r < 1 then

 

[LCKurtz]

In the case of r = -1, the sequence oscillates so it is divergent.

In the second part $y=\ln(|r|^n) = n \ln(|r|)$ goes to infinity whenever r > 1, when 0 < r < 1 then it converges.

So the main behavior to try and examine occurs between -1 < r < 1 then

Yes, that's what I'm getting at. What happens to $y$ when $|r|<1$?

 

[trap101]

Yes, that's what I'm getting at. What happens to $y$ when $|r|<1$?

In that instance it appears as if it's going to converge to 0, but when r is in the form $ln(r)$ it's tending toward negative infiniti

 

[LCKurtz]

In that instance it appears as if it's going to converge to 0, but when r is in the form $ln(r)$ it's tending toward negative infiniti

You can't just assert $|r|^n\to 0$; that's what you are trying to prove if $|r|<1$. But once you know $\ln|r|^n\to -\infty$, doesn't that show it?

 

[trap101]

But once you know $\ln|r|^n\to -\infty$, doesn't that show it?

After staring and thinking about it for a second is this the proper rationale behind the convergence to 0?:

SO I know: $\ln|r|^n\to -\infty$ .

if I raise both sides to "e": e ^{$ln|r|^n$} --> e^$-\infty$. since that would simplify to |r|ⁿ --> 1/ e^$infty$ and that tends to 0.

 

[trap101]

But once you know $\ln|r|^n\to -\infty$, doesn't that show it?

After staring and thinking about it for a second is this the proper rationale behind the convergence to 0?:

SO I know: $\ln|r|^n\to -\infty$ .

if I raise both sides to "e": e ^{$ln|r|^n$} --> e^$-\infty$. since that would simplify to |r|ⁿ --> 1/ e^$infty$ and that tends to 0.

Is that how it ends up converging to 0?

 

[LCKurtz]

After staring and thinking about it for a second is this the proper rationale behind the convergence to 0?:

SO I know: $\ln|r|^n\to -\infty$ .

if I raise both sides to "e": e ^{$ln|r|^n$} --> e^$-\infty$. since that would simplify to |r|ⁿ --> 1/ e^$infty$ and that tends to 0.

That's the idea, although you can probably find a more proper way to express it.

 

[trap101]

Thanks. and thanks for having the patience to help me through it