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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 $b_{2n-1}\le b_{2n}$.2012-10-14
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    I see my mistake. I actually tested both b_2n-1=2012-10-14
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    For induction, would I suppose b_2n-1=2012-10-14
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    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.2012-10-14
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    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...2012-10-14
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    It is indeed the first one. You will probably need to apply the definition several times to get what you want.2012-10-14

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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?2012-10-14
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    @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 $n-1=2k-1$.2012-10-14
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    I see, thanks for your help!2012-10-14
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    @Alti: You’re welcome!2012-10-14