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.

Thanks misan! Could you tell me from where(which book) can I find about Herschfeld's Convergence Theorem?

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!