recursively defined numbers, need to show they are nondecreasingby math_question Tags: nondecreasing, recursive 

#1
Apr2909, 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_{n1}+ \cdots +a_1)} + \frac{1}{(a_n + a_{n1}+ \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 nondecreasing, 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 nondecreasing up to a_n, show that [tex] a_{n+1} \geq a_{n} [/tex]. 


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