Show that ln(1 + 1/x) > 1/(1 + x) for x > 0.
Solution
We have the well-known ey > 1 + y for all y not equal to 1. [Prove, for example, by differentiating.] So ln(1 + y) < y. Put y = -1/(1 + x), so 1 + y = x/(1 + x). Hence 1/(1 + x) < - ln( x/(1+x) ) = ln(1 + 1/x).
© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002