math_question
Apr29-09, 12:32 PM
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}} =
\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 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 \,}
H_n = \sum_{k=1}^{n}\frac{1}{k} \text{\, is the $n^{\text{th}}$
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} .
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 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 \,}
H_n = \sum_{k=1}^{n}\frac{1}{k} \text{\, is the $n^{\text{th}}$
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} .