Let fn(x) = ∑0n xi/i! . Show that fn(n) > en/2. [Assume ex - fn(x) = 1/n! ∫0x (x - t)n et dt, and ∫0∞ tn e-t dt = n! ]
Solution
Using the two relations in the square brackets, we find almost immediately that it suffices to show that ∫0n tn e-t dt < ∫n∞ tn e-t dt.
We notice that the integrand (which we will call g(t) ) is positive for all real t and has a maximum at t = n (differentiate). It would certainly be sufficient to show that g(n + x) > g(n - x) for x in the interval (0, n). This turns out to be true. Taking logs, we need to show that n ln(n + x) - x > n ln(n - x) + x. Putting h(x) = n ln(n + x) - n ln(n - x) - 2x, we need to show that h(x) > 0. Differentiating, h'(x) = 2x2 /(n2 - x2) > 0 for x in (0, n). But h(0) = 0, which gives the result.
Comment. The given facts make this a routine computation. It might have been better to delete the facts in [ ] or to require proof that lim f(n) e-n = 1/2.
© John Scholes
jscholes@kalva.demon.co.uk
18 Aug 2001
Last corrected/updated 2 Dec 02