Author Topic: Is this a sequence?  (Read 771 times)

caballero

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Re: Is this a sequence?
« Reply #15 on: December 18, 2022, 09:05:42 PM »
Quote from: L'Hôpital's Rule
lim (x->infinite)(f(x)/g(x)) = lim (x->infinite) (f'(x)/g'(x))
https://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule

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Sum (n=1,infinite)(A*an+B*bn) = A*Sum(an) + B*Sum(bn)
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If Xn = Yn + Zn, if Yn and Zn are convergents, then Xn is convergent; if Yn or Zn ar not convergent, then Xn neither.

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If Sum(an) is convergent (is a real scalar) => lim(n->infinite) (an) = 0
Thus
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If lim(n->infinite) (an) != 0  => Sum(an) is not convergent

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lim (n->infinite)(B/n) = 0 whenever B!= 0, if B=0 then it is undeterminated.


Xn = (A+n)/(B/n-1) = [A / (B/n - 1)] + [n/(B/n-1)]. Let say Yn = [A / (B/n - 1)], Zn = [n/(B/n-1)]
lim (n->infinite)(Yn) = A / (0-1) = -A
lim (n->infinite)(Zn) = infinite / (0-1) = -infinite


Other way, multiplying up and down by "n":
Yn = An/(B-n), Zn = n^2/(B-n).

[Yn], f(n)=An, g(n)=B-n. f'(n) = A, g'(n)=-1. Hence, by the L'Hôpital's Rule, lim Yn = A/-1 = -A
[Zn], f(n)=n^2, g(n)=B-n, f'(n)=2n, g'(n)=-1. Hence, by the L'Hôpital's Rule, lim Zn = lim 2n/-1 = -infinite

None of Yn and Zn are convergents, so Xn neither.
The logic of the error is hidden among the most unexpected lines of the program

mineiro

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Re: Is this a sequence?
« Reply #16 on: January 03, 2023, 07:58:51 AM »
You can check Wolfram site, please, copy whole line and paste:
https://www.wolframalpha.com/input?i2d=true&i=Divide[A%2Bn%2CDivide[B%2Cn]-1]
https://www.wolframalpha.com/input?i2d=true&i=Divide%5Bn*A%2BPower%5Bn%2C2%5D%2CB-n%5D
I'd rather be this ambulant metamorphosis than to have that old opinion about everything

mabdelouahab

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Re: Is this a sequence?
« Reply #17 on: January 03, 2023, 06:27:13 PM »
 :thumbsup:

caballero

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Re: Is this a sequence?
« Reply #18 on: January 03, 2023, 08:41:19 PM »
Interesting site  :thumbsup:. It seems that whatever you think about, anyone has already done it and uploaded to internet  :biggrin:.

This serie

The logic of the error is hidden among the most unexpected lines of the program

mabdelouahab

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Re: Is this a sequence?
« Reply #19 on: January 04, 2023, 03:01:32 AM »
Interesting site  :thumbsup:. It seems that whatever you think about, anyone has already done it and uploaded to internet  :biggrin:.
Could it be automated processing?

caballero

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Re: Is this a sequence?
« Reply #20 on: January 04, 2023, 04:26:56 AM »
> Could it be automated processing?
Don't understand. What you send clicking in the above link is the formula you want calculate. The page receives that and calculates it.
The logic of the error is hidden among the most unexpected lines of the program