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Radius of convergence

  1. Apr 6, 2016 #1
    1. The problem statement, all variables and given/known data

    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

    2. Relevant equations

    Root test, ratio test

    3. 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 gonna 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.

  2. jcsd
  3. Apr 6, 2016 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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.
  4. Apr 6, 2016 #3
    Cool thanks. Yeah for a) I quickly realised 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 |an+1/an| = 1, but now I think that's not it.

    Thank you very much for your answer. :)

  5. Apr 6, 2016 #4


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    Science Advisor

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

    |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|.

    The ratio test will not work for many numerical series. For example, [tex]\sum_{n=0}^\infty n^3[/tex] does not converge but the ratio test gives [tex]\frac{(n+ 1)^3}{n^3}[/tex] 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.
    Last edited by a moderator: Apr 6, 2016
  6. Apr 6, 2016 #5
    @HallsofIvy Thank you that was a very clear explanation.

  7. Apr 6, 2016 #6


    Staff: Mentor

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