In the following solution, for variable $i\in \{ p,q,r\}$, we write $f(i)\prec ,\succ g(i)$ iff there exist integer $C$ that $f(i)<, >g(i)$ for every $i\geq C$. Note that for each $k=2,3,4,...,10$, the condition $\{ \sqrt[k]{n}\} =0.2015...$ is equivalent to $\Big( p+\frac{2015}{10000}\Big)^k<n<\Big( p+\frac{2016}{10000}\Big)^k$ for some positive integer $p$. It's enough to show the following claim for each $k=10,9,8,...,3$; for every sufficiently large $p$, there is an integer $q$ that $I_1\subset I_2$ where $I_1$ and $I_2$ denote intervals $\Big( \Big( q+\frac{2015}{10000}\Big)^{k-1},\Big( q+\frac{2016}{10000}\Big)^{k-1} \Big)$ and $\Big( \Big( p+\frac{2015}{10000}\Big)^k,\Big( p+\frac{2016}{10000}\Big)^k \Big)$. Note that after proving this, we need to choose large enough interval so that there's at least one integer. The rest of this solution is the proof of that claim. Suppose to the contrary that there exist arbitrary large $p$ that the claim is not true. For those positive integer $p$, let $r$ be the greatest positive integer that $\Big( r+\frac{2016}{10000}\Big)^{k-1}<\Big( p+\frac{2015}{10000}\Big)^k$. If $\Big( r+1+\frac{2015}{10000}\Big)^{k-1}>\Big( p+\frac{2016}{10000}\Big)^k$, we get that $\Big( r+1+\frac{2015}{10000}\Big)^{k-1}-\Big( r+\frac{2016}{10000}\Big)^{k-1}>\Big( p+\frac{2016}{10000}\Big)^k-\Big( p+\frac{2015}{10000}\Big)^k$. We have $(k-1)r^{k-2}\succ LHS>RHS>\frac{k}{10000}p^{k-1}\Rightarrow r\succ c\times p^{\frac{k-1}{k-2}}$ for a positive constant $c$. Recall the inequality condition of $r$, we have $r<\Big( p+\frac{2015}{10000}\Big)^{\frac{k}{k-1}}<(p+1)^{\frac{k}{k-1}}$. So $c\times p^{\frac{k-1}{k-2}} \prec (p+1)^{\frac{k}{k-1}}$. It's equivalent to $c\times p^{\frac{1}{(k-1)(k-2)}}\times \frac{p}{p+1} \prec \Big( 1+\frac{1}{p}\Big)^{\frac{1}{k-1}}$. By Bernoulli's, $RHS<1+\frac{1}{p(k-1)}$, so $c\times p^{\frac{1}{(k-1)(k-2)}}\times \frac{p}{p+1} \prec 1+\frac{1}{p(k-1)}$, clearly false for large enough $p$. So $\Big( r+1+\frac{2015}{10000}\Big)^{k-1}<\Big( p+\frac{2016}{10000}\Big)^k$. We have two possible cases, but by definition of $r$, both give us there are two intervals which are intersecting and one of the endpoints of an interval is $\Big( r+1+\frac{2015}{10000}\Big)^{k-1}$. Consider the endpoint, called it $s$, of another interval that lies in $\Big( \Big( p+\frac{2015}{10000}\Big)^k,\Big( p+\frac{2016}{10000}\Big)^k \Big)$. If we can prove that from both sides from $s$, one of distance toward an endpoint is greater than $\Big( r+2+\frac{2015}{10000}\Big)^{k-1}-\Big( r+1+\frac{2015}{10000}\Big)^{k-1}$, that side would surely contain one whole interval we wanted. Note that such distance $\prec k\Big( r+1+\frac{2015}{10000}\Big)^{k-2}$ So it's enough to show that $\Big( p+\frac{2016}{10000}\Big)^k-\Big( p+\frac{2015}{10000}\Big)^k\succ 2\times k\Big( r+1+\frac{2015}{10000}\Big)^{k-2} $. We have $LHS>\frac{k}{10000}p^{k-1}$, so we need $p^{k-1}\succ 20000\times \Big( r+1+\frac{2015}{10000}\Big)^{k-2} $ Recall the condition of $r$, we get $\Big( r+1+\frac{2015}{10000}\Big)^{k-2} <\Big( (p+1)^{\frac{k}{k-1}}+1+\frac{2015}{10000}\Big)^{k-2} \prec p^{k-1}$. The last inequality is true by Binomial's. This completes our claim and so completes the problem.