Cauchy's Integral Test for Convergence

  • #1
27
0
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.
 

Answers and Replies

  • #2
334
47
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
22,089
3,296
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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