# I Cauchy's Integral Test for Convergence

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1. Jun 20, 2016

### samgrace

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. Jun 21, 2016

### Incand

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

3. Jun 22, 2016

### chiro

4. Jun 23, 2016

### micromass

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