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I Cauchy's Integral Test for Convergence

  1. Jun 20, 2016 #1
    Hello,

    I am want to prove that: $$ \sum_{1}^{\infty} \frac{1}{n^{2} + 1} < \frac{1}{2} + \frac{1}{4}\pi $$

    Cauchy's Convergence Integral

    If a function decreases as n tends to get large, say f(x), we can obtain decreasing functions of x, such that:

    $$ f(\nu - 1) \geqslant f(x) \geqslant f(\nu) $$

    where

    $$ {\nu - 1 \leqslant x \leqslant \nu} $$

    we then let:

    $$ V_{\nu} = f(\nu-1) - \int^{\nu}_{\nu-1} f(x) .dx = \int^{\nu}_{\nu-1} f(\nu-1) - f(x) .dx $$

    to give the values of the constraints;

    $$ 0 \leqslant V_{\nu} \leqslant f(\nu-1) - f(\nu) $$

    My understanding of this is that V sub nu is a value of the limit for each evaluated interval, thus it would be less than the evaluation itself. The next step is

    We take the sum of V sub nu, and this should have the constraint;

    $$ V_{2} + V_{3} + ... + V_{n} \leqslant f(1) - f(n) \leqslant f(1) $$

    Which is in my attempt, then the final part which I think I'm not understand correctly:

    V sub nu is then convergent to a positive limit as n -> infinity:

    $$ \sum_{1}^{n-1} V_{\nu} - \int^{n}_{1} f(x) .dx \leqslant f(1)$$

    thus as the function decreases as n tends to get large:

    $$ f(1) + f(2) + ... + f(n) $$

    It may or may not converge, just as:

    $$ \int^{n}_{1} f(x) .dx $$

    does or does not tend to the limit l as n --> infinity.

    then

    Of which I can't get the equality displayed at the start.


    My attempt

    With values for nu:

    $$ \nu = 1, 2, 3, 4, 5 $$

    with the x between the integer increment to be half the interval:

    $$ x = 0.5, 1.5, 2.5, 3.5, 4.5. $$

    I obtained values for the limit of each interval evaluation:

    $$ V_{1} = 0.2 \\ V_{2}= 0.19 \\ V_{3} = 0.06 \\ V_{4} = 0.02 \\ V_{5} = 0.011 $$

    Then the constraint:

    $$ V_{2} + V_{3} + V_{n} \leqslant f(1) - f(n) \leqslant f(1) $$

    holds as;

    $$ 0.281 \leqslant 0.4615 \leqslant 0.5 $$

    So the intergrand taken for n=x=5:

    $$ \sum_{1}^{n-1} V_{\nu} - \int^{n}_{1} f(x) .dx $$

    equals:

    $$ 0.281 - 0.153 = 0.128 $$

    Where

    $$ \frac{1}{2} + \frac{1}{4}\pi = 1.128 $$

    Can anyone tell me where the one comes in? I thought it was the the truncated series of V sub nu.

    It works if you add in the f(0) term:

    $$ f(0) + \sum_{1}^{n-1} V_{\nu} - \int^{n}_{1} f(x) .dx \leqslant f(1) $$

    $$ 1 + 0.281 - 0.153 = 1.128 $$


    I hope this makes sense and someone can clarify the limits to me.
     
  2. jcsd
  3. Jun 21, 2016 #2
    Since you haven't gotten that many replies so far I thought I add another way of showing the inequality. I can't help you with the method you're using.

    ##\sum_1^\infty \frac{1}{n^2+1} \le -\frac{1}{2}+\sum_1^\infty \frac{1}{n^2}##
    where the last sum is well known https://en.wikipedia.org/wiki/Basel_problem
     
  4. Jun 22, 2016 #3

    chiro

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  5. Jun 23, 2016 #4

    micromass

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    The exact answer is ##\frac{\pi}{2}\text{coth}(\pi)## by the way. That doesn't help, but it's neat to know.
     
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