Is this a sequence?

[caballero]

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

Sum (n=1,infinite)(A*a_n+B*b_n) = A*Sum(a_n) + B*Sum(b_n)

If Xn = Yn + Zn, if Yn and Zn are convergents, then Xn is convergent; if Yn or Zn ar not convergent, then Xn neither.

If Sum(an) is convergent (is a real scalar) => lim(n->infinite) (an) = 0

Thus

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

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.

[mineiro]

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

[mabdelouahab]

 

[caballero]

Interesting site  . It seems that whatever you think about, anyone has already done it and uploaded to internet  .

This serie

[mabdelouahab]

Interesting site  . It seems that whatever you think about, anyone has already done it and uploaded to internet  .

Could it be automated processing?

[caballero]

> 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.