Show that for any integer n > 1, we have 1/e - 1/(ne) < (1 - 1/n)n < 1/e - 1/(2ne).
Take the left-hand inequality first. Multiplying through by e and taking logs, we want to show that 0 < 1 + (n-1) ln(1 - 1/n). Putting x = 1/n, it is sufficient to show that ln(1 - x) > x/(1-x) for 0 < x < 1. But we can use the expansions: ln(1 - x) = -(x + x2/2 + x3/3 + ... ), and x/(1-x) = x + x2 + x3 + ... and the result is immediate.
Similarly, for the right-hand inequality we need: 1 + n ln(1 - 1/n) < ln(1 - 1/2n). Putting x = 1/n, it is sufficient to show that -(x/2 + x2/3 + x3/4 + ... ) < -(x/2 + (x/2)2/2 + (x/2)3/3 + ... ) for 0 < x < 1. Again this holds term by term because n2n > n+1 for n > 1.
63rd Putnam 2002
© John Scholes
11 Dec 2002