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

by math_question
Tags: non-decreasing, recursive
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math_question
#1
Apr29-09, 11:32 AM
P: 2
Hello, I have a question involving a sequence of numbers [tex] \{a_n\} [/tex] defined recursively. They are defined as the positive solutions of the following set of equations.

[tex]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 [/tex]

The first few can be solved analytically [tex] a_1 = 1, a_2 \approx 1.62, \cdots [/tex]. 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. [tex] a_{n+1} \geq a_{n} [/tex] 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. ([tex]\ln{(2n+2)}[/tex] seems to fit the numerical solution perfectly. ) Also, [tex]a_n \geq H_n, \text{\, where \,}
H_n = \sum_{k=1}^{n}\frac{1}{k} \text{\, is the $n^{\text{th}}$
harmonic number.} [/tex]

The numerical solution suggests [tex] a_{n+1} - a_n [/tex] 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 [tex] a_{n+1} \geq a_{n} [/tex].
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