Radius of Convergence for Power Series: What is the Limiting Ratio Test?

[Telemachus]

Homework Statement

Hi there. Well, I was trying to determine the radius and interval of convergence for this power series:
$$\displaystyle\sum_{0}^{\infty} \displaystyle\frac{x^n}{n-2}$$

So this is what I did till now:
$$\displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{n-1}{n-2}}\right |}=1$$

Then $$\left |{x}\right |<1$$
The thing is this result is clearly wrong, I think the series diverges for any x, there is a discontinuity at n=2 for the sum. But I don't know what I'm doing wrong.

 

[LCKurtz]

Homework Statement

Hi there. Well, I was trying to determine the radius and interval of convergence for this power series:
$$\displaystyle\sum_{0}^{\infty} \displaystyle\frac{x^n}{n-2}$$

I would hazard a guess that the n-2 in the denominator is a typo or else this problem appeared on some quiz to see if you were awake. But, ignoring the first three terms...

So this is what I did till now:
$$\displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{n-1}{n-2}}\right |}=1$$

Doesn't your ratio test involve an x?

Then $$\left |{x}\right |<1$$

But there was no x in your limit above, so where did you get that?

The thing is this result is clearly wrong, I think the series diverges for any x, there is a discontinuity at n=2 for the sum. But I don't know what I'm doing wrong.

It does diverge if the n-2 thing is correct.

 

[Telemachus]

Thanks LCKurtz. It is correct the n-2 in the denominator. Anyway, I've considered that n=/=2.

This is how I've reasoned it at first:
If exists $$\displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n+1}(x-x_0)^{n+1}}{a_{n}(x-x_0)^{n}}}\right |}<1$$

Then
$$R=\displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n}}{a_{n+1}}}\right |}<(x-x_0)$$

 

[eczeno]

hi,

i have never seen a series like that, very interesting. i can't think of any way to call that a convergent series with that rouge term in there, not even at the center. your analysis of the radius of convergence is correct for the remaining terms.

cheers

 

[HallsofIvy]

The question is not just the "n-2" in the front but the n= 0 as the starting n for the sum. Then the sum is for n= 0, n= 1, n= 2- and the summand does not exist for n= 2?

 

[LCKurtz]

Thanks LCKurtz. It is correct the n-2 in the denominator. Anyway, I've considered that n=/=2.

This is how I've reasoned it at first:
If exists $$\displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n+1}(x-x_0)^{n+1}}{a_{n}(x-x_0)^{n}}}\right |}<1$$

Then
$$R=\displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n}}{a_{n+1}}}\right |}<(x-x_0)$$

OK, here we are assuming the sum starts at 3. In determining the radius of convergence, you don't have any choice about x₀. This is a power series expanded about 0. And you should write your ratio like this:

$$\left| \frac{a_{n+1}}{a_n}\right|=\left|\frac{(n-1)}{(n-2)}\frac{x^{n+1}}{x^n} \right| = \left| \frac {(n-1)}{(n-2)}\right| |x|$$

When you let n → ∞ you just get |x|, and for convergence, you want that limit ratio to be less than 1. That's where you get |x| < 1 for your interval of convergence.

 

[Telemachus]

The question is not just the "n-2" in the front but the n= 0 as the starting n for the sum. Then the sum is for n= 0, n= 1, n= 2- and the summand does not exist for n= 2?

Right, I've assumed it's not defined for n=2, and proceeded as well. If I take n=2 as part of the sum, I have to consider it diverges I think, but I'm not sure, its not defined for n=2. It would be a contradiction to consider that it diverges, now I see, if considered n=2 as part of the sum, because of that theorem that says that for a power series there always exists an interval of convergence.

OK, here we are assuming the sum starts at 3. In determining the radius of convergence, you don't have any choice about x₀. This is a power series expanded about 0. And you should write your ratio like this:

$$\left| \frac{a_{n+1}}{a_n}\right|=\left|\frac{(n-1)}{(n-2)}\frac{x^{n+1}}{x^n} \right| = \left| \frac {(n-1)}{(n-2)}\right| |x|$$

When you let n → ∞ you just get |x|, and for convergence, you want that limit ratio to be less than 1. That's where you get |x| < 1 for your interval of convergence.

Its the same thing expressed in different ways. I do it this way because its how my professor gave it in class, but the result is the same. I've just "passed" one part of the equation to the other side of the inequality and called that as the radius of convergence.

Perhaps the summation starting at n=0 was just a misspelling.

 

[LCKurtz]

Thanks LCKurtz. It is correct the n-2 in the denominator. Anyway, I've considered that n=/=2.

This is how I've reasoned it at first:
If exists $$\displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n+1}(x-x_0)^{n+1}}{a_{n}(x-x_0)^{n}}}\right |}<1$$

Then
$$R=\displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n}}{a_{n+1}}}\right |}<(x-x_0)$$

Its the same thing expressed in different ways. I do it this way because its how my professor gave it in class, but the result is the same. I've just "passed" one part of the equation to the other side of the inequality and called that as the radius of convergence.

Well, maybe it's just a typo, but it certainly isn't correct. That last inequality is backwards and needs absolute value signs on the right.

 

[Telemachus]

It isn't backwards :P it follows from here:
$$\displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n+1}(x-x_0)^{n+1}}{a_{n}(x-x_0)^{n}}}\right |}<1 \rightarrow \displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n+1}(x-x_0)}{a_{n}}}\right |}<1 \rightarrow \displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n+1}}{a_{n}}}\right |}< \displaystyle\frac{1}{(x-x_0)}$$

Then: $$R = \displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n}}{a_{n+1}}}\right |}<(x-x_0)$$

 

[LCKurtz]

It isn't backwards :P it follows from here:
$$\displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n+1}(x-x_0)^{n+1}}{a_{n}(x-x_0)^{n}}}\right |}<1 \rightarrow \displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n+1}(x-x_0)}{a_{n}}}\right |}<1 \rightarrow \displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n+1}}{a_{n}}}\right |}< \displaystyle\frac{1}{(x-x_0)}$$

Then: $$R = \displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n}}{a_{n+1}}}\right |}<(x-x_0)$$

You can not invert both sides of an inequality without reversing the inequality sign and it still need absolute values on the right, even though it's wrong.

 

[Telemachus]

Sorry, yes, the inequality sign goes on the opposite sense. Missed that, and the absulute values on the x-x_0
This is it:
$$\displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n+1}(x-x_0)^{n+1}}{a_{n}(x-x_0)^{n}}}\right |}<1 \rightarrow \displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n+1}(x-x_0)}{a_{n}}}\right |}<1 \rightarrow \displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n+1}}{a_{n}}}\right |}< \displaystyle\frac{1}{|x-x_0|} \rightarrow R = \displaystyle\lim_{n \to{+}\infty}{\left |{\displaystyle\frac{a_{n}}{a_{n+1}}}\right |}>|x-x_0|$$