Let x2 = 21/2, x3 = (2 + 31/3)1/2, x4 = (2 + (3 + 41/4)1/3)1/2, ... , xn = (2 + (3 + ... + n1/n ... )1/3)1/2 (where the positive root is taken in every case). Show that xn+1 - xn < 1/n! .
Solution
Solution by Demetres Christofides
We have (a1/n - b1/n)(a(n-1)/n + a(n-2)/nb1/n + ... + a1/nb(n-2)/n + b(n-1)/n) = a - b. So if a > b ≥ 1, then the long parenthesis is ≥ n and hence (a1/n - b1/n) ≤ (a - b)/n.
Applying this repeatedly, we get xn+1 - xn ≤ ( (3 + ... + (n+1)1/(n+1))1/3 - (3 + ... + n1/n)1/3 )/2! ≤ ( (4 + ... + (n+1)1/(n+1))1/4 - (4 + ... + n1/n)1/4)/3! ≤ ... ≤ (n+1)1/n+1 1/n!. So that almost gets us there, but not quite.
Note that (n + 1/n)n < 3, so (n + 1)n < 3.nn ≤ nn+1 for n ≥ 3. Taking the n(n+1)th root, we get (n + 1)1/(n+1) < n1/n for n ≥ 3. Also 41/4 = 21/2, so (n + 1)1/(n+1) ≤ 21/2 for n ≥ 3.
We now make the first inequality in the chain slightly stronger. We have a1/2 - b1/2 = (a - b)/(a1/2 + b1/2) < (a - b)/(21/2 + 21/2) = (a - b)/(2 21/2) for a > b ≥ 2. So the chain gives xn+1 - xn < (n+1)1/(n+1) (1/n!) (1/21/2) ≤ 1/n! for n ≥ 3.
It remains to show that x3 - x2 < 1/2. We have x3 - x2 = 31/3/( (2 + 31/3)1/2 + 21/2) < 1.5/( 31/2 + 21/2) < 1.5/3 < 1/2.
© John Scholes
jscholes@kalva.demon.co.uk
11 Sep 2002