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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 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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    @Alti: You’re welcome!2012-10-14