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Homework Help: Proving the convergence of series

  1. Nov 26, 2017 #1
    1. The problem statement, all variables and given/known data
    Prove the convergence of this series using the Comparison Test/Limiting Comparison Test with the geometric series or p-series. The series is:

    The sum of [(n+1)(3^n) / (2^(2n))] from n=1 to positive ∞

    The question is also attached as a .png file

    2. Relevant equations

    The geometric series with a * r^n is known to:
    • converge if the absolute value of r is smaller than 1
    • diverge if the absolute value of r is greater or equal to 1
    The p-series (1/n^p) is known to:
    • converge if p is greater than 1
    • divergent if otherwise
    Please refer to this website for the definition of the Comparison Test and the Limiting Comparison Test:

    3. The attempt at a solution
    This is as far as I got:
    ∑ [(n+1)(3^n) / 2^(2n)] can be split into two series,
    ∑ [n(3^n) / (4^n)] + ∑ [(3^n) / (4^n)]
    The latter series is known to converge because it is a geometric series with r = 3/4
    However, I am stuck trying to solve the first series.

    Using the Ratio Test to determine whether the series converges or diverges is quite simple, and I have worked it out.

    By the way, could anyone tell me how to make the formulas look more ... natural? Instead of using ^ and / as well as a ton of brackets.

    Any help is appreciated!
     

    Attached Files:

  2. jcsd
  3. Nov 26, 2017 #2

    fresh_42

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    Can you trap ##(n+1)(\dfrac{3}{4})^n## and a geometric series ##\sum q^n\,##?
     
  4. Nov 26, 2017 #3
    Could you elaborate? I don't quite understand what you mean by ∑qn
     
  5. Nov 26, 2017 #4

    fresh_42

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    What is a geometric series? You're requested to use one. So you need to know what it is, which of them converge and finally what comparison means. If you can answer all three questions correctly, you're almost there.
     
  6. Nov 26, 2017 #5

    Mark44

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    The first series can be written as ##\sum n (3/4)^n##. What can you say about this series by using the Comparison Test or Limit Comparison Test?
     
  7. Nov 26, 2017 #6
    This is exactly what I am having trouble solving. I could try to move n into the brackets with 3/4.
    ∑n(3/4)n = ∑(n√n)n(3/4)n = ∑(n√n 3/4)n
    Then I am stuck.
     
  8. Nov 26, 2017 #7

    fresh_42

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    You have to find a quotient ##q## with ##(n+1)\left( \dfrac{3}{4} \right)^n < q^n##. Any ideas how to find a ##q\,##? Hint: find a quotient ##p## with ##(n+1) < p^n## first.
     
  9. Nov 26, 2017 #8
    I am having trouble finding a p that satisfies pn>(n+1)
    :cry:
     
  10. Nov 26, 2017 #9
    In this case the quotient would need to be ... greater than 1?
     
  11. Nov 26, 2017 #10

    fresh_42

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    Well, ##p=2## would certainly do, for all ##n>2##.
    Exactly. Now in the example with ##p=2## we will lose too much space, because we need ##p^n \cdot \left( \dfrac{3}{4} \right)^n < q^n < 1##.
    This narrows it down, but there are still enough possibilities.
     
  12. Nov 26, 2017 #11
    What about p = 11/10 ? pn would be greater than (n+1) if n is sufficiently large, and pn * (3/4)n = (33/40)n which would satisfy r < 1.
     
  13. Nov 26, 2017 #12

    fresh_42

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    Yes, that would do. You still need to prove, that it gets larger than ##n+1## (binomial formula) for which (?) ##n## and sweep everything together for a formal proof.
     
  14. Nov 26, 2017 #13
    I could set up the equation (11/10)n = n+1 but I would not be able to solve it without the aid of a graphing calculator. Using a graphing calculator, I found that for all x > 40, pn is greater than n+1. Thus, I would have proved that when n goes from 1 to positive infinity, the series (33/40)n > (n+1)(3/4)n.

    Could you suggest a simpler method? In my Calculus 2 course, the use of the binomial formula was not mentioned.

    Thanks!
     
  15. Nov 26, 2017 #14

    fresh_42

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    Well, you have to explain why ##n+1 < \left(\dfrac{11}{10} \right)^n ## for large ##n\;[*]##. I find it easiest by using ##\left(\dfrac{11}{10} \right)^n = \left(1+\left( \dfrac{1}{10}\right) \right)^n=1+\binom{n}{1}\left(\dfrac{1}{10} \right)+\binom{n}{2}\left(\dfrac{1}{10} \right)^2+ \ldots + \left(\dfrac{1}{10} \right)^{11}## but you can as well proof that ##x \longmapsto (x+1)^{\frac{1}{x}}## is monotone decreasing with limit ##1##, or by induction that ##(n+1)^{\frac{1}{n}} > (n+2)^{\frac{1}{n+1}}## or rule out by another method, that it cannot increase for large ##n## at some point. Maybe it's even enough to simply say, that exponential behavior outnumbers polynomial behavior, or just take it for granted - I don't know. For a rigorous proof, we need ##(n+1) \cdot \left( \dfrac{3}{4} \right)^n {<}_{[*]} \left(\dfrac{11}{10} \right)^n \cdot \left( \dfrac{3}{4} \right)^n = \left( \dfrac{33}{40} \right)^n## and the value of ##\sum_{n \in \mathbb{N}} \left( \dfrac{33}{40} \right)^n##. Thus some argument for ##[*]## is needed, in my opinion.
     
    Last edited: Nov 26, 2017
  16. Nov 26, 2017 #15

    Ray Vickson

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    Have you had the fact that
    $$\lim_{x \to \infty} \frac{\ln (x)}{x} = 0 ?$$
    If so, that means that for any fixed positive constant ##c = \ln(1+\alpha) ## (with ##\alpha > 0##) there is some ##N_{\alpha}## such that ##\ln(x) < c x## for ##x > N_{\alpha}##, hence ##n < e^{nc} = (1+\alpha)^n ## for ##n > N_{\alpha}##. Therefore, for ##0 < r < 1## we have ##n r^n < (r(1+\alpha))^n## for ##n > N_{\alpha}##. As long as you choose ##\alpha## so that ##(1+\alpha) r \equiv \rho < 1## your terms ##n r^n## are bounded above by a convergent geometric sequence ##\rho^n## for all ##n## beyond a certain point. For given ##r## and ##\rho \in (r,1)## you can figure out what values of ##\alpha## will work, then pick one such value and thus figure out what is ##N_{\alpha}##. However, for just proving the result you do not need to do any numerics; it is enough to know that certain things exist (and, in principle, could be found explicitly if need be).
     
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