- #1
samgrace
- 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.
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.