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Let $b_1=1$ and $b_n=1+\frac{1}{1+b_{n-1}}$

Prove that $(b_{2k-1})$ is increasing for $k \in \mathbb{N}$

By definition, a sequence $(a_{n})$ is increasing if $a_{n}≤a_{n+1}$ for all $n \in \mathbb{N}$.

SO, for this problem, must prove $b_{2n-1}≤b_{2n}$ for all $n$.

Proceed by induction:

Start with $n=1$. Then, $b_1=1$ and $b_2=3/2$, so $b_1≤b_2$.

Assume inductively that $b_{2n-1}≤b_{2n}$, prove $b_{2n}≤b_{2n+1}$.

Am I doing this correctly? I want to know before I continue.

Thanks.

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It's always useful to write out at least a few terms of the sequence. Your sequence of convergents beings with $$\left(1,\ \frac{3}{2},\ \frac{7}{5}, \frac{17}{12},\ \frac{41}{29},\ \cdots\right)$$ You can see that the sequence itself is not increasing. The even terms form a decreasing sequence and the odd terms form the increasing sequence. You need to prove that $b_{2n-1}\le b_{2(n+1)-1} = b_{2n+1}$ and not $b_{2n-1}\le b_{2n}$. –  EuYu Oct 14 '12 at 2:33
    
I see my mistake. I actually tested both b_2n-1=<b_2n and b_2n-1=<b_2n+1 but I guess I made some algebraic mistake. Thanks for the clarification, i'll fix it. –  Alti Oct 14 '12 at 2:35
    
For induction, would I suppose b_2n-1=<b_2n+1 and then prove b_2n+1=<b_2n+3? –  Alti Oct 14 '12 at 2:43
    
That's certainly a viable approach. Of course, you can never be sure before hand what will work and what won't, but that's the first thing I would try anyways. –  EuYu Oct 14 '12 at 2:44
    
Okay, and I know this might be obvious, but just wanted to make sure, is b_2n+1=1+[1/(1+b_2n)] or b_2n+1=1+[1/(1+b_2n-1)]. I think it's the first one, but I want to make sure... –  Alti Oct 14 '12 at 2:51
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1 Answer

up vote 0 down vote accepted

Here is a non-inductive approach. Since $$b_n=1+\frac{1}{1+b_{n-1}}\;,$$

the double step from $b_{n-1}$ to $b_{n+1}$ is given by

$$b_{n+1}=1+\frac1{1+b_n}=1+\frac1{2+\frac1{1+b_{n-1}}}=1+\frac1{\frac{3+2b_{n-1}}{1+b_{n-1}}}=1+\frac{1+b_{n-1}}{3+2b_{n-1}}=\frac{4+3b_{n-1}}{3+2b_{n-1}}\;.$$

Then $$b_{n+1}-b_{n-1}=\frac{4+3b_{n-1}}{3+2b_{n-1}}-b_{n-1}=\frac{4-2b_{n-1}^2}{3+2b_{n-1}}=\frac{2(2-b_{n-1}^2)}{3+2b_{n-1}}\;,$$ which is positive if and only if $2-b_{n-1}^2>0$, i.e., if and only if $b_{n-1}^2<2$. Now use the earlier problem.

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Thank you. For b_{n-1}^2<2, I showed that b_{2k-1}^2<2, will it still yield the same results? –  Alti Oct 14 '12 at 20:32
    
@Alti: What you showed earlier was that $b_{n-1}^2<2$ if $n-1$ is odd, which is exactly what you need here. If $n-1$ is odd, then $n$ is even, so $n=2k$ for some integer $k$, and $n-1=2k-1$. –  Brian M. Scott Oct 14 '12 at 22:52
    
I see, thanks for your help! –  Alti Oct 14 '12 at 23:12
    
@Alti: You’re welcome! –  Brian M. Scott Oct 14 '12 at 23:14
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