Showing $2^{n_2} + 3^{n_3}+\cdots+9^{n_9}$ is dense in $\mathbb{R}^+$ I encountered this problem via a friend. He asked me to prove that 
$$ \left\{u: u= \sum_{k=2}^9 k^{n_k} \quad n_k \in \mathbb{Z}  \right\}$$ 
is dense in $\mathbb{R}^+$.
I was able to show that $0$ is approachable(can get arbitrarily close to) by numbers of this type. However to proceed ahead, I need to know what operations preserve the structure of these numbers. For instance, to show $m+n\sqrt{2}$ is dense for all integers $m,n$, I used the fact that product of any two numbers in this form yields a number in the same form. My problem is I don't know what pattern to exploit in the given problem.
I appreciate any hints / patterns you can provide. One idea I have been working on is as follows:
If we could partition $2^{n_2}, 3^{n_3},\ldots, 9^{n_9}$ into two parts, one that would determine the integer part, the other that would approximate the fractional part, we would be done. 
 A: I don't believe that this is true.
First consider the part of the sum involving negative exponents.  The most that this part can be is
$$2^{-1}+\cdots+9^{-1}<\frac{1}{2}+\Bigl(\frac{1}{3}+\frac{1}{6}\Bigr)+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}=2\ .$$
Now consider the parts with non-negative exponents.  Let $n$ be a positive integer.  For $k=2,\ldots,9$, the number of powers of $k$ in the range from $1$ to $10^n$ is
$$1+\left\lfloor n\frac{\log10}{\log k}\right\rfloor<5n\ .$$
Therefore the number of sums of powers of $2,\ldots,9$ with non-negative exponents, the sum not exceeding $10^n$, is at most $(5n)^8$.  For large $n$ this is very much less than $10^n$; so there will in some cases be gaps of at least $3$ between attainable sums; and from the first part of the argument, these gaps cannot be "bridged" by taking some terms with negative exponents.
A: I don't think this can be done. Think of it like this: Suppose you have $u \in \Bbb{R}_+$ and $u_n \to u$ with
$$ u_n = \sum_{k=2}^9 k^{m_k^n} $$
Now each $(u_n)$ is completely characterized by $(m_2^n,...,m_9^n)$. If $u$ is not zero then not all sequences $(m_k^n)_n$ go to $-\infty$, so some of them are bounded, which means that we will have some values repeated an infinity of times. Then we can extract a diagonal sequence such that every $(m_k^n)_n$ which does not go to $-\infty$ is constant. This means that you approximate $u=2^{a_2}+...+9^{a_9}$ where $a_1,...,a_9$ are integers or $-\infty$ (with convention $a^{-\infty}=0$ for $a>1$.) 
Obviously not every $u$ can be written in this form.
In the same way you can prove that the colsure of the set you mention is precisely
$$ \{ u =2^{a_1}+...+9^{a_9} : a_i \in \Bbb{Z} \cup \{-\infty\}\} $$
