lim (x->infinite)(f(x)/g(x)) = lim (x->infinite) (f'(x)/g'(x))

https://en.wikipedia.org/wiki/L'H%C3%B4pital's_ruleSum (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.