The sequence ${a_{n}}$ $n\in \mathbb{N}$ is given in a recursive way with $a_{1}=1$, $a_{n}=\prod_{i=1}^{n-1} a_{i}+1$, for all $n\geq 2$. Determine the least number $M$, such that $\sum_{n=1}^{m} \frac{1}{a_{n}} <M$ for all $m\in \mathbb{N}$
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Tags: number theory
RagvaloD
16.07.2017 15:42
Old problem $a_{n+1}=1+\prod_{i=1}^{n} a_{i}=1+a_n (\prod_{i=1}^{n-1} a_{i})=1+a_n(a_n-1)=a_n^2-a_n+1$ Let $b_m=\sum_{n=1}^{m} \frac{1}{a_{n}} $ $b_1=1,b_2=1+\frac{1}{2}=2-\frac{1}{2}=2-\frac{1}{a_3-1},b_3=1+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}=2-\frac{1}{6}=2-\frac{1}{a_4-1}$ We will prove, that $b_{n}=2-\frac{1}{a_{n+1}-1}$ Let it is true for $n-1$ then $b_n=b_{n-1}+\frac{1}{a_n}=2-\frac{1}{a_n-1}+\frac{1}{a_n}=2-\frac{1}{a_n(a_n-1)}=2-\frac{1}{a_{n+1}-1}$ So $b_{n}=2-\frac{1}{a_{n+1}-1}$ and $\lim b_n =2$ Answer : $M=2$
WolfusA
14.08.2019 21:26
41. Deutsche Mathematik-Olympiade 2001/2002 (final round in Germany). There was modified version: $a_1$ is a fixed positive real number, then $$\sum_{n=1}^{m} \frac{1}{a_{n}} <\frac{2}{a_1}$$and that's the best upper bound that holds for all $m\in Z_+$.