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A sequence $a_{1},a_{2},a_{3},\cdots, $ is defined by $$a_{n}=2a_{n}a_{n+1}+3a_{n+1}$$ for all $n=1,2,3,\cdots,$ if $b_{n}=1+\dfrac{1}{a_{n}}$ for all $n=1,2,\cdots$ find the largest integer $m$ such that $$\sum_{k=1}^{n}\dfrac{1}{n+\log_{3}{b_{k}}}>\dfrac{m}{24}$$ for all positive integer $n\ge 2$

my idea:we have $$\dfrac{1}{a_{n+1}}+1=3(1+\dfrac{1}{a_{n}})$$ so $$b_{n+1}=3b_{n}$$ then we have $$ b_{n}=b_{1}\cdot 3^{n-1}$$ let $\log_{3}(1+\dfrac{1}{a_{1}})=c$

then we $$\sum_{k=1}^{n}\dfrac{1}{n+\log_{3}{b_{k}}}=\dfrac{1}{n+c}+\dfrac{1}{n+1+c}+\dfrac{1}{n+2+c}+\cdots+\dfrac{1}{2n-1+c}$$

so $$m<24\left(\dfrac{1}{n+c}+\dfrac{1}{n+1+c}+\dfrac{1}{n+2+c}+\cdots+\dfrac{1}{2n-1+c}\right)_{min}$$ so I think $$24\left(\dfrac{1}{n+c}+\dfrac{1}{n+1+c}+\dfrac{1}{n+2+c}+\cdots+\dfrac{1}{2n-1+c}\right)_{min} =24\lim_{n\to\infty}\left(\dfrac{1}{n+c}+\dfrac{1}{n+1+c}+\dfrac{1}{n+2+c}+\cdots+\dfrac{1}{2n-1+c}\right)=24\ln{2}=16.635$$ so $m=16$?

but this problem answer $m=13$

meaning is my wrong? Thank you

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you haven't given a starting value $a_1$. Was this intentional? –  john Jun 7 '13 at 15:21
    
Thank you,@john This problem isn't given the starting value $a_{1}$ –  math110 Jun 7 '13 at 15:23

1 Answer 1

You've done all the hard work to show that $$\sum_{k=1}^n\frac1{n+\log_3b_k} =\sum_{k=1}^n\frac1{n+k+c}.$$ If this is $>\frac m{24}$ for all $n\ge 2$, then especially $$\frac1{3+c}+\frac1{4+c}>\frac m{24}.$$ For the claimed answer $m=13$ to be correct, a quick function plot shows that this requires that $c\in (-4,-3.5665\ldots)\cup(-3, 0.25881\ldots)$, i.e. $a_1\in(-\infty,-1.0384\ldots)\cup(-1.02027\ldots,-1.0125)\cup(3.0406\ldots,\infty)$. So unless you are given more information about $a_1$, we cannot even verify if $m=13$ is a good answer or not. At least you see from this that

  • Not only the limit decides about the minimum
  • The precise answer depends on $a_1$
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