Cauchy's Integral Test for Convergence

In summary, Cauchy's convergence integral can be used to find a function that decreases as n tends to get large. This can be used to find a decreasing function of x, and then the constraints can be used to find the limit of the function.
  • #1

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:

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


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

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.


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:

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


$$ 0.281 - 0.153 = 0.128 $$


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

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.
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  • #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
  • #4
The exact answer is ##\frac{\pi}{2}\text{coth}(\pi)## by the way. That doesn't help, but it's neat to know.

What is Cauchy's Integral Test for Convergence?

Cauchy's Integral Test is a method used to determine the convergence or divergence of an infinite series. It is based on the use of integrals and was developed by mathematician Augustin-Louis Cauchy.

How does Cauchy's Integral Test work?

Cauchy's Integral Test states that if the integral of the function f(x) from 1 to infinity exists, then the infinite series ∑ f(n) converges. If the integral diverges, then the series also diverges.

What are the conditions for using Cauchy's Integral Test?

The function f(x) must be continuous, positive, and decreasing on the interval [1, infinity] for Cauchy's Integral Test to be applicable. Additionally, the integral of f(x) from 1 to infinity must be solvable.

What is the advantage of using Cauchy's Integral Test?

Cauchy's Integral Test can be easier to use than other convergence tests, such as the Ratio Test or the Comparison Test. It also allows for the use of integrals, which can be more familiar and easier to work with for some mathematicians.

Are there any limitations to Cauchy's Integral Test?

Cauchy's Integral Test can only be used to determine convergence or divergence, not to find the sum of a series. It also may not be applicable to all types of series, such as alternating series or series with terms that do not decrease monotonically.

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