Understanding Radius of Convergence in Power Series Calculations

[JulienB]

Homework Statement

Hi everybody! I'm a little struggling to fully understand the idea of radius of convergence of a function, can somebody help me a little? Are some examples I found in old exams at my university:
Calculate the radius of convergence of the following power series:
a) ∑(x-1)^j/(2^j + 3^j)
b) ∑(1 + j + j²)⋅x^j
c) ∑(2^j + 3^j)⋅x^j

Homework Equations

Root test, ratio test

The Attempt at a Solution

So I found a solution for each of them, but I don't know if it's right and I will include a few questions I have here and there:

Let's first see if I get the "methods" right:
I want to try first the ratio test because the computation is easier. And here is my first question: if lim sup j→∞ |a_j+1/a_j| = 1, I cannot conclude anything and I should instead run the root test, is that right? Or is the convergence radius R = 1 anyway?

For a), I run the ratio test and get:

a_j+1/a_j = (2^j + 3^j)/(2^j+1 + 3^j+1) = 1/3

That means that R = 3 and that ∑(x-1)^j/(2^j + 3^j) absolutely converges for |x-1| < 3 ⇔ |x| < 4. Is that correct or am I completely missing the point? What does it really mean that the power series converges absolutely for |x| < 4?

b) This time I'm going to try the root test:

(1 + j + j²)^1/j = (j²(1 + (1/j) + (1/j²)))^1/j
= (1 + (1/j) + (1/j))^1/j

I notice that 1 ≤ 1 + (1/j) + (1/j) ≤ 2 for any j ≥ 2, therefore

1^1/j ≤ (1 + (1/j) + (1/j))^1/j ≤ 2^1/j
1^1/j → 1
2^1/j → 1
⇒ (1 + (1/j) + (1/j))^1/j → 1

Therefore R = 1 and ∑(1 + j + j²)⋅x^j absolutely converges for |x| < 1.

c) It's very similar to a) so I run the root test this time:

(2^j + 3^j)^1/j = (3^j)^1/j⋅(1 + (2/3)^j)^1/j
= 3⋅(1 + (2/3)^j)^1/j → 3 because:

1^1/j ≤ (1 + (2/3)^j)^1/j ≤ 2^1/j

Therefore R = 1/3 and ∑(2^j + 3^j)⋅x^j absolutely converges for |x| ≤ 1/3.

Are those right? Any remarks about the topic?

Thank you very much in advance for your help.Julien.

 

[Ray Vickson]

Homework Statement

Hi everybody! I'm a little struggling to fully understand the idea of radius of convergence of a function, can somebody help me a little? Are some examples I found in old exams at my university:
Calculate the radius of convergence of the following power series:
a) ∑(x-1)^j/(2^j + 3^j)
b) ∑(1 + j + j²)⋅x^j
c) ∑(2^j + 3^j)⋅x^j

Homework Equations

Root test, ratio test

The Attempt at a Solution

So I found a solution for each of them, but I don't know if it's right and I will include a few questions I have here and there:

Let's first see if I get the "methods" right:
I want to try first the ratio test because the computation is easier. And here is my first question: if lim sup j→∞ |a_j+1/a_j| = 1, I cannot conclude anything and I should instead run the root test, is that right? Or is the convergence radius R = 1 anyway?

For a), I run the ratio test and get:

a_j+1/a_j = (2^j + 3^j)/(2^j+1 + 3^j+1) = 1/3

That means that R = 3 and that ∑(x-1)^j/(2^j + 3^j) absolutely converges for |x-1| < 3 ⇔ |x| < 4. Is that correct or am I completely missing the point? What does it really mean that the power series converges absolutely for |x| < 4?

b) This time I'm going to try the root test:

(1 + j + j²)^1/j = (j²(1 + (1/j) + (1/j²)))^1/j
= (1 + (1/j) + (1/j))^1/j

I notice that 1 ≤ 1 + (1/j) + (1/j) ≤ 2 for any j ≥ 2, therefore

1^1/j ≤ (1 + (1/j) + (1/j))^1/j ≤ 2^1/j
1^1/j → 1
2^1/j → 1
⇒ (1 + (1/j) + (1/j))^1/j → 1

Therefore R = 1 and ∑(1 + j + j²)⋅x^j absolutely converges for |x| < 1.

c) It's very similar to a) so I run the root test this time:

(2^j + 3^j)^1/j = (3^j)^1/j⋅(1 + (2/3)^j)^1/j
= 3⋅(1 + (2/3)^j)^1/j → 3 because:

1^1/j ≤ (1 + (2/3)^j)^1/j ≤ 2^1/j

Therefore R = 1/3 and ∑(2^j + 3^j)⋅x^j absolutely converges for |x| ≤ 1/3.

Are those right? Any remarks about the topic?

Thank you very much in advance for your help.Julien.

For (a): the statement $|x-1| < 3$ is correct, but the statement $|x| < 4$ is wrong.
In (b) the ratio test would also work, and would be easier.
(c) is OK.

 

[JulienB]

For (a): the statement $|x-1| < 3$ is correct, but the statement $|x| < 4$ is wrong.
In (b) the ratio test would also work, and would be easier.
(c) is OK.

Cool thanks. Yeah for a) I quickly realized my mistake that if x is negative, then it would be wrong. Should I just let |x - 1| < 3 then unless the problem specifically asks for |x|?
For b) I used the root test just to have some variety. About the ratio test, it says on Wikipedia that it does not work in many cases but is vague about which: would you happen to know when? At first I was under the impression that it was when |a_n+1/a_n| = 1, but now I think that's not it.Thank you very much for your answer. :)Julien.

 

[HallsofIvy]

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I want to try first the ratio test because the computation is easier. And here is my first question: if lim sup j→∞ |aj+1/aj| = 1, I cannot conclude anything and I should instead run the root test, is that right? Or is the convergence radius R = 1 anyway?

Remember that your series also had an (x- 1)^j term. So your ratio is NOT just \left|\frac{a_{n+1}}{a_n}\right|, it is \left|\frac{a_{n+1}}{a_n}\right|\frac{|x- 1|^{j+1}}{|x- 1|^j}= \left|\frac{a_{n+1}}{a_n}\right||x- 1|. The power series converges absolutely if and only if that is less than 1 so if and only if |x- 1|&lt; \frac{1}{\left|\frac{a_{n+1}}{a_n}\right|}= \frac{1}{R}.

Should I just let |x - 1| < 3 then unless the problem specifically asks for |x|?

|x- 1|< 3 is the same as -3< x- 1< 3. Adding 1 to each part, -2< x< 5. That cannot be written in terms of |x|.

About the ratio test, it says on Wikipedia that it does not work in many cases but is vague about which: would you happen to know when?

The ratio test will not work for many numerical series. For example, \sum_{n=0}^\infty n^3 does not converge but the ratio test gives \frac{(n+ 1)^3}{n^3} which goes to 1 so the ratio test does not work. For a power series, there will always be some values of x that make it less than 1.

 

[JulienB]

@HallsofIvy Thank you that was a very clear explanation.Julien.

 

[Mark44]

|x- 1|< 3 is the same as -3< x- 1< 3. Adding 1 to each part, -2< x< 5.

The last inequality should be -2 < x < 4. Geometrically, the inequality |x - 1| < 3 represents the numbers that are within 3 units of 1.