recursively defined numbers, need to show they are non-decreasing

**math_question** · Apr29-09, 12:32 PM

Hello, I have a question involving a sequence of numbers $LaTeX Code: \\{a_n\\}$ defined recursively. They are defined as the positive solutions of the following set of equations.

$LaTeX Code: 1 = \\frac{1}{a_1} = \\frac{1}{(a_2+a_1)} + \\frac{1}{{a_2}} = \\frac{1}{(a_3 + a_2 +a_1)} + \\frac{1}{(a_3 + a_2)} + \\frac{1}{{a_3}} = \\cdots = \\frac{1}{(a_n + a_{n-1}+ \\cdots +a_1)} + \\frac{1}{(a_n + a_{n-1}+ \\cdots +a_2)} + \\cdots + \\frac{1}{{a_n}} = \\cdots$

The first few can be solved analytically $LaTeX Code: a_1 = 1, a_2 \\approx 1.62, \\cdots$ . The rest can be found numerically.

However, instead of their values, what I need is to show that these numbers have to be non-decreasing, i.e. $LaTeX Code: a_{n+1} \\geq a_{n}$ for all n.

Any ideas welcome.

Thank you.

Notes:

An observation that might be helpful is that, the numerical solutions show a_n's grow with natural logarithm. ( $LaTeX Code: \\ln{(2n+2)}$ seems to fit the numerical solution perfectly. ) Also, $LaTeX Code: a_n \\geq H_n, \\text{\\, where \\,} H_n = \\sum_{k=1}^{n}\\frac{1}{k} \\text{\\, is the $n^{\\text{th}}$ harmonic number.}$

The numerical solution suggests $LaTeX Code: a_{n+1} - a_n$ goes to zero as n goes to infinity. So this might suggest induction is the way to go for the proof. Assume non-decreasing up to a_n, show that $LaTeX Code: a_{n+1} \\geq a_{n}$ .