Wouter asked
f[x_] := Sum[(-1)^(n-j) Binomial[j,n-j] x^(n-j),{j,1,n}] / Sum[(-1)^(n+2-j) Binomial[j,n+2-j] x^(n+2-j),{j,1,n+2}] produces a well known Generating Function for the limit n->Infinity.
Q: What is needed to force Limit[f[x], n->Infinity] into submission? it equals (1-2x - Sqrt[1 - 4*x]) / (2 x^2)
Mathematica's package Calculus`Limit` doesn't feel like tackling it. Should it?
and I blathered
Both Mma and Macsyma need a little help here. Just send the sum limits to infinity first. It might also help to start from j=0, which you may have intended. If not, Mma may need you to "manually" subtract the j=0 term. Then you can even start from j=-inf. The resulting sums are simple algebraics(n), and n->inf is a snap.
Bull. This isn't even a limit problem. But it exposes several defects in my personally enhanced Macsyma. All that should have been necessary was (declare(n,integer), hypersimp(sum(...))). Instead, I have to sum from floor(n/2) instead of 1: (c225) (declare(n,integer),sum((-1)^(n-j)*binomial(j,n-j)*x^(n-j),j,floor(n/2),n) /sum((-1)^(n+2-j)*binomial(j,n+2-j)*x^(n+2-j),j,floor(n/2),n+2)) n ==== \ n + j n - j > binomial(j, n - j) (- 1) x / ==== n j = floor(-) 2 (d225) --------------------------------------------------------- n + 2 ==== \ n + j n - j + 2 > binomial(j, n - j + 2) (- 1) x / ==== n j = floor(-) 2 then manually reverse the summation direction, (c226) rangereverse(%) n floor(-) - 1 2 ==== \ n + j n - j > binomial(j, n - j) (- 1) x / ==== j = n + 1 (d226) --------------------------------------------------------- n floor(-) - 1 2 ==== \ n + j n - j + 2 > binomial(j, n - j + 2) (- 1) x / ==== j = n + 3 and then manually change the summation variable: (c227) linchange(%,n+1-j) n n - floor(-) 2 ==== j \ binomial(n - j, j) x > (- ---------------------) / j ==== (- 1) j = 0 (d227) ---------------------------------------------- n n - floor(-) 2 ==== j + 2 \ binomial(n - j, j + 2) x > (- -----------------------------) / j ==== (- 1) j = - 2 Finally, (c228) hypersimp(%) 4 (d228) -------------------- 2 (sqrt(1 - 4 x) + 1) I assume this is the square of what you'd have gotten with f(n)/f(n+1) instead of f(n)/f(n+2). --rwg