Asymptotic Formula for (1 + 1/x)x, Revisited

In a recent note [1] appearing in this Monthly, Chen and Choi calculated a j in 1 + 1 x x = e ∞ j=0 a j x j (1) We offer a briefer calculation of these same coefficients. Applying the change of variables y = 1 x , and the two functions f (y) = e y , g(y) = ln(1+y) y , we rewrite (1) as f (g(y)) = ∞ j=0 (ea j)y j We take derivatives of this formal power series j times, and substitute y = 0, to get d j dy j f (g(y))| y=0 = e(j!)a j (2) The left side of (2) may be calculated withFà a di Bruno's famous formula, which gives d j dy j f (g(y)) = j! k 1 ! · · · k j ! f (k 1 +···+k j) (g(y)) g (1) (y) 1! k 1 · · · g (j) (y) j! k j where the sum is taken over all solutions to k 1 + 2k 2 + · · · + jk j = j. Because g(y) = 1 − y 2 + y 2 3 − y 3 4 + · · · , we have g (t) (y)| y=0 = (−1) t t! t+1. We also have lim y→0 f (k 1 +···+k j) (g(y)) = lim y→0 f (g(y)) = e. Combining, we get e(j!) 1 k 1 ! · · · k j ! 1 2 k 1 · · · 1 j + 1 k j (−1) k 1 +2k 2 +···+jk j = e(j!)a j We now cancel e(j!) from both sides, and use the k 1 + 2k 2 + · · · + jk j = j restriction , to get the main result from [1]: