### Putnam 1996

**Problem B2**

Let f(n) = ( (2n+1)/e )^{(2n+1)/2}. Show that for n > 0: f(n-1) < 1.3.5 ... (2n-1) < f(n).

**Solution**

*Straightforward integration, with one wrinkle.*

We estimate the integral of ln x, which is convex and hence easy to estimate. Take the integral from 1 to 2n-1. This is less than 2(ln 3 + ln 5 + ... + ln(2n-1) ). But the integral of ln x is x ln x - x, so the integral evaluates to (2n-1) ln(2n-1) - 2n + 2. Hence (2n-1) ln(2n-1) - (2n-1) < (2n-1) ln(2n-1) - 2n + 2 < 2(ln 3 + ln 5 + ... + ln(2n-1)). Exponentiating gives the left hand inequality.

Similarly, the integral from 1 to 2n+1 is greater than 2(ln 3 + ln 5 + ... + ln(2n-1) ). However, this is not quite good enough, because the integral evaluates to (2n+1) ln(2n+1) - 2n, and this time the odd 1 works against us. However, we can slightly improve our estimate. Because ln 1 = 0, our estimate treated the integral from 1 to 3 as nil, so we can raise the lower end of the integration range: the integral from 3 to 2n+1 is greater than 2(ln 3 + ln 5 + ... + ln(2n-1) ). Actually, it is more convenient to take the lower bound as e < 3, because then x ln x - x evaluates to 0 at the lower bound, thus getting rid of the odd 1. Exponentiating now gives the right hand inequality.

Putnam 1996

© John Scholes

jscholes@kalva.demon.co.uk

12 Dec 1998