TITLE: Showing $2^{n_2} + 3^{n_3}+\cdots+9^{n_9}$ is dense in $\mathbb{R}^+$
QUESTION [6 upvotes]: 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.

REPLY [6 votes]: 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.