- #1
JulienB
- 408
- 12
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/(2j + 3j)
b) ∑(1 + j + j2)⋅xj
c) ∑(2j + 3j)⋅xj
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→∞ |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?
For a), I run the ratio test and get:
aj+1/aj = (2j + 3j)/(2j+1 + 3j+1) = 1/3
That means that R = 3 and that ∑(x-1)j/(2j + 3j) 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
11/j ≤ (1 + (1/j) + (1/j))1/j ≤ 21/j
11/j → 1
21/j → 1
⇒ (1 + (1/j) + (1/j))1/j → 1
Therefore R = 1 and ∑(1 + j + j2)⋅xj absolutely converges for |x| < 1.
c) It's very similar to a) so I run the root test this time:
(2j + 3j)1/j = (3j)1/j⋅(1 + (2/3)j)1/j
= 3⋅(1 + (2/3)j)1/j → 3 because:
11/j ≤ (1 + (2/3)j)1/j ≤ 21/j
Therefore R = 1/3 and ∑(2j + 3j)⋅xj absolutely converges for |x| ≤ 1/3.
Are those right? Any remarks about the topic?
Thank you very much in advance for your help.Julien.