a) Let n1,n2,… be a sequence of natural number such that ni≥2 and ϵ1,ϵ2,… be a sequence such that ϵi∈{1,2}. Prove that the sequence: n1√ϵ1+n2√ϵ2+⋯+nk√ϵkis convergent and its limit is in (1,2]. Define n1√ϵ1+n2√ϵ2+… to be this limit. b) Prove that for each x∈(1,2] there exist sequences n1,n2,⋯∈N and ni≥2 and ϵ1,ϵ2,…, such that ni≥2 and ϵi∈{1,2}, and x=n1√ϵ1+n2√ϵ2+…
Problem
Source: Iranian National Olympiad (3rd Round) 2007
Tags: induction, limit, real analysis, real analysis unsolved
28.08.2007 12:27
29.08.2007 08:13
Omid Hatami wrote: a) Let n_{1},n_{2},\dots be a sequence of natural number such that n_{i}\geq2 and \epsilon_{1},\epsilon_{2},\dots be a sequence such that \epsilon_{i}\in\{1,2\}. Prove that the sequence: \sqrt[n_{1}]{\epsilon_{1}+\sqrt[n_{2}]{\epsilon_{2}+\dots+\sqrt[n_{k}]{\epsilon_{k}}}} is convergent and its limit is in (1,2] i *think* the result for a) will follow once you apply Herschfeld's Convergence Theorem, which states --- for some p_{k}\in\mathbb R,\;\;\;0<p_{k}<1 and \epsilon_{k}>0 ( k\in\mathbb Z,\;\;\;k>1), ............. \lim_{k\to\infty}\left(\epsilon_{1}+\left(\epsilon_{2}+\left(\epsilon_{3}+\left(\epsilon_{4}+\dots+\left(\epsilon_{k}\right)^{p_{k}}\right)^{p_{4}}\right)^{p_{3}}\right)^{p_{2}}\right)^{p_{1}} converges iff (\epsilon_{i})^{(p_{i})^{i}} is bounded for some i\in\mathbb Z,\;\;\;i>1. in your example, \epsilon_{i}\in\{1,2\} is bounded, hence the result should follow immediately. in other words, first try and prove the sequence in question is bounded. and then if you can show it monotonically increases or decreases, then you can prove that the sequence is convergent.
29.08.2007 16:42
Thanks misan! Could you tell me from where(which book) can I find about Herschfeld's Convergence Theorem?
29.08.2007 16:56
b) is a little dirtier. We will construct a sequence of n_{1},n_{2}... and e_{1},e_{2}...\in\{1,2\} for an arbitrary x. First observe that \bigcup_{n = 2}^{\infty}(\sqrt [n]{2},\sqrt [n]{4}] = (1,2](*)(supposing not we will get a contradiction like 2^{n+1}\ge4^{n} for an n\ge 2 which is obviously false) We construct inductively: for 1 we choose n_{1} so that we have \sqrt [n_{1}]{1+1}< x and \sqrt [n_{1}]{1+2}\ge x and we take e_{1}= 1 or \sqrt [n_{1}]{2+1}< x and \sqrt [n_{1}]{2+2}\ge x and we take e_{1}= 2. using(*) this n_{1} exists, because the condition is the existence of an n so that x\in(\sqrt [n]{2},\sqrt [n]{4}]. For 2 we choose n_{2} so that we have \sqrt [n_{1}]{e_{1}+\sqrt [n_{2}]{1+1}}< x and \sqrt [n_{1}]{e_{1}+\sqrt [n_{2}]{1+2}}\ge x and we take e_{2}= 1 or \sqrt [n_{1}]{e_{1}+\sqrt [n_{2}]{2+1}}< x and \sqrt [n_{1}]{e_{1}+\sqrt [n_{2}]{2+2}}\ge x and we take e_{2}= 2. Using (*) and the way we have chosen n_{1} this n_{2} exists. \vdots for k we can choose n_{k} and e_{k} so that we have \sqrt [n_{1}]{e_{1}+\sqrt [n_{2}]{e_{2}+...+\sqrt [n_{k}]{e_{k}+1}}}< x and \sqrt [n_{1}]{e_{1}+\sqrt [n_{2}]{e_{2}+...+\sqrt [n_{k}]{e_{k}+2}}}\ge x So denote a_{k}=\sqrt [n_{1}]{e_{1}+\sqrt [n_{2}]{e_{2}+...+\sqrt [n_{k}]{e_{k}}}} We will prove that this converges to x. First obviously a_{k}<\sqrt [n_{1}]{e_{1}+\sqrt [n_{2}]{e_{2}+...+\sqrt [n_{k}]{e_{k}+1}}}< x Then denote t_{k}= x-a_{k}> 0. Then t_{k}\le\sqrt [n_{1}]{e_{1}+\sqrt [n_{2}]{e_{2}+...+\sqrt [n_{k}]{e_{k}+2}}}-a_{k} So \sqrt [n_{1}]{e_{1}+\sqrt [n_{2}]{e_{2}+...+\sqrt [n_{k}]{e_{k}}}}+t_{k}\le\sqrt [n_{1}]{e_{1}+\sqrt [n_{2}]{e_{2}+...+\sqrt [n_{k}]{e_{k}+2}}} Raise both sides to n_{1}th power. We have( t_{k} positive): e_{1}+\sqrt [n_{2}]{e_{2}+...+\sqrt [n_{k}]{e_{k}}}+n_{1}a_{k}^{n-1}{t_{k}\le (\sqrt [n_{1}]{e_{1}+\sqrt [n_{2}]{e_{2}+...+\sqrt [n_{k}]{e_{k}}}}+t_{k})^{n_{1}}\le} \le e_{1}+\sqrt [n_{2}]{e_{2}+...+\sqrt [n_{k}]{e_{k}+2}}, so ( a_{k}> 1) we have \sqrt [n_{2}]{e_{2}+...+\sqrt [n_{k}]{e_{k}}}+n_{1}t_{k}<\sqrt [n_{2}]{e_{2}+...+\sqrt [n_{k}]{e_{k}+2}}. We repeat this procedure and finally we will have: n_{1}n_{2}...n_{k}t_{k}< 2 from here( n_{i}\ge 2) we have that t_{k}\rightarrow 0 so a_{k}\rightarrow x. Hope there aren't any mistakes. Beautiful problem!