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

[math_question]

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

1 = \frac{1}{a_1} = \frac{1}{(a_2+a_1)} + \frac{1}{{a_2}} =<br /> \frac{1}{(a_3 + a_2 +a_1)} + \frac{1}{(a_3 + a_2)}<br /> + \frac{1}{{a_3}} = \cdots = \frac{1}{(a_n + a_{n-1}+ \cdots +a_1)} + \frac{1}{(a_n + a_{n-1}+<br /> \cdots +a_2)} + \cdots + \frac{1}{{a_n}} = \cdots

The first few can be solved analytically 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. 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. (\ln{(2n+2)} seems to fit the numerical solution perfectly. ) Also, a_n \geq H_n, \text{\, where \,}<br /> H_n = \sum_{k=1}^{n}\frac{1}{k} \text{\, is the $n^{\text{th}}$<br /> harmonic number.}

The numerical solution suggests 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 a_{n+1} \geq a_{n}.

 

[fresh_42]

$a_2=\frac{1-\sqrt{5}}{2} \approx -0.62$ is also a solution, which means $a_1=1 < a_2\,.$
Hence you cannot prove it without additional conditions. Let us assume the positive root $a_2=\frac{1+\sqrt{5}}{2}$. Then I get for $a_3$ again two negative roots and one positive. So without a criterion how to select the positive solution instead of a negative one, this cannot be proven from the recursion alone. If we added $a_n>0$ as an assumption, then we first would have to show that there is always such a solution. And the degrees of the $n-$th term is $n$, so as you said, it can only be solved numerically. Maybe this can help: