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Convergence of the summation

  1. Jan 7, 2017 #1
    1. The problem statement, all variables and given/known data
    I'm give the following summation of functions and I have to see where it converge.
    $$\sum_{n = 1}^{\infty} \frac{(3 arcsin x)^n}{\pi^{n + 1}(\sqrt(n^2 + 1) + n^2 + 5)}$$

    2. Relevant equations


    3. The attempt at a solution
    Putting ##3 arcsin x = y##, I already see that with the theorem of D'Alambert I have a range ##r = \pi##.
    The summation then converges in the interval:
    $$|3 arcsin x| < \pi$$
    which is
    $$sin(-\frac{\pi}{3}) < x < sin(\frac{\pi}{3})$$
    Now I don't know it the summation converges in the two external points, so I try with the first one and I get:
    $$\frac{1}{\pi}\sum_{n = 1}^{\infty} \frac{(-1)^n}{(\sqrt(n^2 + 1) + n^2 + 5)}$$
    In this case, I decide to use the Leibniz criterion. So first I find if the summation converges to ##0## and then I find if decreases monotonically.
    Doing the limit of the summation (minus the ##(-1)^n##) I get that it converges to ##0##.
    Now, to find out if it decreases monotonically, I do the first derivative of the following:
    $$f(n) = \frac{1}{(\sqrt(n^2 + 1) + n^2 + 5)}$$
    and I end up with
    $$f'(n) = -\frac{\frac{x}{\sqrt(x^2 + 1)} + 2x}{(\sqrt(n^2 + 1) + n^2 + 5)^2}$$
    Now I have to put this ##>0## but due to the minus sign it will be ##<0##. The one at the denominator is always positive, so we have to look at only the numerator. We will have then:
    $$\frac{x}{\sqrt(x^2 + 1)} + 2x < 0$$
    But, strangely, I don't get how to continue from here.
    Can someone help me?
     
  2. jcsd
  3. Jan 7, 2017 #2

    fresh_42

    Staff: Mentor

    ##x \mapsto \frac{x}{\sqrt{x^2+1}}+2x## is positive for ##x>0##, negative for ##x<0## and zero at ##x=0##.
     
  4. Jan 7, 2017 #3

    haruspex

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    I do not understand how differentiating f(n) wrt n caused x to appear from nowhere.
    Since 1/n2 converges, it is rather obvious that the sum you have converges at both ends of the range. Is there not a test along those lines?
     
  5. Jan 7, 2017 #4

    Ray Vickson

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    You have ##f(n) = 1/d(n)##, where ##d(n) = \sqrt{n^2+1} + n^2+5## is a strictly increasing function of ##n##. No calculus is needed here: ##n^2## is increasing in ##n##, so ##\sqrt{n^2+1}## is increasing, as is ##n^2+5##. You are just adding two increasing functions.
     
  6. Jan 8, 2017 #5
    Why? How did you do that?

    Sorry, my bad! I meant "n" where there was "x".

    So it decreases monotonically? This means that both this and the other one for ##\frac{\pi}{3}## converge and so it all converges to the interval ##[-\frac{\pi}{3} , \frac{\pi}{3}]##?
     
  7. Jan 8, 2017 #6

    Ray Vickson

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    Why do you assume you have convergence at both end points?
     
  8. Jan 8, 2017 #7

    fresh_42

    Staff: Mentor

    ##\frac{x}{\sqrt{x^2+1}}+2x=x \cdot \underbrace{(2+\frac{1}{+\sqrt{x^2+1}})}_{>\,2}##
     
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