Fisrt of all A must be infinite. Assume that the density of A in $\mathbb{N}$ is well defined. Denote by $p_{i}$ the probability that an element of A lie in the set $b_{i}A+c_{i}$. This probability is equal to \[p_{i}= \lim_{n\to \infty}\frac{\#\{b_{i}A+c_{i}\cap A \}\cap \{1,2,..n\}}{\#A \cap \{1,2,..n\}}=\] \[\lim_{n\to \infty}\frac{\#\{x\in A, b_{i}x+c_{i}\le n \}}{\#A \cap \{1,2,..n\}}=\] \[\lim_{n\to \infty}\frac{\#A \cap \{1,2,..\frac{n-c_{i}}{b_{i}}\}}{\#A \cap \{1,2,..n\}}\le \frac 1{b_{i}}\] Since one of this events may not occur and they exclude each other we have that $\sum \frac 1{b_{i}}\le 1$. Now we only need to prove that the density of A in $\mathbb{N}$ is well defind i.e. $\lim_{n\to infty}d_{n}=\lim_{n\to \infty}\frac{\#A\cap \{1,..,n\}}{n}$ exists. Can anybody fill in this argument?

I put the official solution below,but I can't understand the step $ \frac{\phi(N+N_{0})}{\phi (N)}\ge\frac{1}{q^{N_{0}}}$ can somebody explain this to me? And I am interested in the solution given in the comment,but the comment didn't give a full solution.I am surprised by the hint of the comment,but the last step which lead to the result didn't have a proof.I don't know if it is trivial or it is just a theorem about infinite set,could anyone continue the proof?

How stupid I am I knew the ineq now. But if we define the density as ZetaX in http://www.mathlinks.ro/viewtopic.php?p=849974#849974 $ d_{A}(B) : =\lim_{k\to\infty}\frac{n_{B}(k)}{n_{A}(k)}$ Can we obtain $ d_{A}(A_{i})=\frac{1}{b_{i}}$?

dear hxy09, using the second approach (the comment in the official solution) leads to the following solution. Lemma. For a set $ A \subset \mathbf{N}$ and $ x \in \mathbf{N}$, let us define $ f(x) = \#\{y\mid y\in A,y\leq x\}$ if $ u_1,\ldots,u_n$ are real numbers such that $ \sum u_i > 1$ then there exists $ x$ such that $ f(u_i \cdot x)\geq u_i \cdot f(x)$ for all $ i = 1,\ldots,n$ This is not very intuitive, but you can prove it using contradiction. (I think it is a wonderful result) the lemma is nontrivial, but as a brief outline, assume contradiction and work on from the large numbers and find a fast increasing sequence of $ \frac {f(x)}{x}$, which is smaller than 1 for all $ x$, and obtain a contradiction. using the lemma one can finish the problem easily.

ywl wrote: dear hxy09, using the second approach (the comment in the official solution) leads to the following solution. Lemma. For a set $ A \subset \mathbf{N}$ and $ x \in \mathbf{N}$, let us define $ f(x) = \#\{y\mid y\in A,y\leq x\}$ if $ u_1,\ldots,u_n$ are real numbers such that $ \sum u_i > 1$ then there exists $ x$ such that $ f(u_i \cdot x)\geq u_i \cdot f(x)$ for all $ i = 1,\ldots,n$ This is not very intuitive, but you can prove it using contradiction. (I think it is a wonderful result) the lemma is nontrivial, but as a brief outline, assume contradiction and work on from the large numbers and find a fast increasing sequence of $ \frac {f(x)}{x}$, which is smaller than 1 for all $ x$, and obtain a contradiction. using the lemma one can finish the problem easily. Thank you very much,but can you give details?

I was trying to find another solution (but can't finish it). Let $ f(x)=\sum_{a\in A}x^a$. Using the conditions in the problem we get $ 1\geq \sum_{i=1}^n x^{c_i}\frac{f(x^{b_i})}{f(x)}$. Can we prove (I don't even know if it's true) that $ \lim_{x\to 1}\frac{f(x^b)}{f(x)}=\frac{1}{b}$?

limes123 wrote: I was trying to find another solution (but can't finish it). Let $ f(x) = \sum_{a\in A}x^a$. Using the conditions in the problem we get $ 1\geq \sum_{i = 1}^n x^{c_i}\frac {f(x^{b_i})}{f(x)}$. Can we prove (I don't even know if it's true) that $ \lim_{x\to 1}\frac {f(x^b)}{f(x)} = \frac {1}{b}$? If $ \lim_{x\to 1}{f(x)(1-x)} =a>0$ then $ \lim_{x\to 1}\frac {f(x^b)}{f(x)} = \lim_{x\to 1}\frac {f(x^b)(1-x^b)}{f(x)(1-x)} \cdot \lim_{x\to 1}\frac {1-x}{1-x^b} = \frac {1}{b}$ so it remains to prove that $ \lim_{x\to 1}{f(x)(1-x)} =a>0$ but I dont know how to prove this. And we can use the condition \[ {1\over b_{1}}+\,\ldots\,+{1\over b_{n}}>1.\].

Hello, I believe I have a different solution that uses nothing on density (just some really bad bounds) Let Y be the sum (1/a) for a in A and a <= (b_n)^j. Clearly Y < 1+1/2+...+1/(b_n)^n Let f_i(x) be defined as b_i*x +c_i Let Z be the sum (1/f(f(f(...x)) for some sequence of j f's, where each f is one of f_i. It is clear Z<Y since the elements in Z's sum are in Y's sum. but f_d(x)f_e(x) > b_db_e(x+2*max(c_i) (this generalizes really easily but is a huge pain to write out) so Z > (sum(1/b_i))^j * 1/(x+j*max(c_i)) where x is a generator of a. if sum(1/b_i) > 1, then let it = x. then we have x^j/j*constant <<<< j*ln(b_n) for all j, since x^j/j<Z<Y<1+1/2+...+1/(b_n)^n ~ j ln (b_n) But this is false for large j. So sum (1/b_i) <=1

Solution by dysfunctionalequations and me.

Maybe this is equivalent to the density argument but it's much easier for me to see... By (1), $A$ is infinite. Note that for all $s>1$, $\sum_{a\in A}\frac{1}{a^s}\le\zeta(s)$ converges. Also, for all $\epsilon>0$, there exists an integer $M$ such that for all $a\in A,a\ge M$ and $1\le i\le n$, $b_ia+c_i\le b_ia(1+\epsilon)$ (since the $b_i,c_i$ are fixed positive integers). By the problem hypothesis, \[\sum_{i=1}^{n}\sum_{a\in A,a\ge M}\frac{1}{b_i^s a^s(1+\epsilon)^s}\le\sum_{i=1}^{n}\sum_{a\in A,a\ge M}\frac{1}{(b_ia+c_i)^s}\le \sum_{a\in A,a\ge M}\frac{1}{a^s},\]i.e. \[\sum_{i=1}^{n}\frac{1}{b_i^s}\le(1+\epsilon)^s\]for all $s>1$ and $\epsilon>0$. The conclusion follows immediately by continuity.

dear @math154 desinty may be good for you, but mayn't for I. Please, who can solve the problem, by Generating function?

Dear mathuz, I don't know why you are not satisfied with the density arguments, but it's the most natural approach. Imagine a perfect situation when the density of $A$ is the same in every interval $[1,N]\,,\, N\in \mathbb{N}$ . Then the density of $b_i A+c_i$ in respect to $A$ (as being a subset of $A$) , framed in the interval $[1,N]$ , slightly differs $\frac{1}{b_i}$ , when $n\to \infty$. It is so, because $A$ has an infinitely many elements and the proportion of elements of $b_i \cdot\{a\in A \mid a\leq N/b_i\} $ , that leave the interval $[1,N]$ after translation by $c_i$ , will be as small as we want as $N\to \infty$. This is the main idea, it only remains to carry out the general case Denote (as above) the density of $A$ in $[0,N]$ as $d_N(A) = |\{a\in A \mid a\leq N\}|/N$. The above idea applies without changes if there are infinitely many $N$ with property $d_N(A) \leq \min_{k<N} d_k(A)$. In this case the density of $\{b_i a + c_i\ \mid a\in A, a\leq N/b_i \}$ in respect to $\{a\in A \mid a\leq N\}$ is bigger than $\frac{1}{b_i} - \varepsilon$ , where $\varepsilon$ can be taken arbitrary small. If it is not the case, then $\liminf_{N\to\infty} d_N(A) = d > 0$. We can take $(N_k)_{k=1}^{\infty}$ with $\lim_{k\to\infty} d_{N_k} (A) = d$ and apply the above arguments to $N=N_k$. It gets a little bit more complicated since involves some routine "epsilon techniques". Here $d>0$ is an important thing. So, the idea is more than clear, the only difficulties are pure technical, although it is not a problem if one has some calculus experience. The Victor's solution above, essentially resembles density argument, but avoids all these annoying details one must care about.

Solution : Consider linear functions $\psi_i :\mathbb{N}\to\mathbb{N}$, where $\psi_i(n) := b_in + c_i$ ($b_i\geq 2$). Define sets $S(n): = \{\psi_{i_1}^{j_1}\psi_{i_1}^{j_1}\ldots \psi_{i_N}^{j_N}(n)|i_1, i_2, \ldots, i_N$, $j_1, j_2, \ldots, j_N\in\mathbb{N}\}$ for every $n\in \mathbb{N}$. $\textbf{Lemma 1:}$ In these assumptions we have that $\sum_{i\in S(n)}\frac{1}{i}<\infty$ for any $n\in\mathbb{N}$. $\textbf{Lemma 2:}$ We have that $\lim_{n\to\infty}\frac{\frac{1}{b_j}\sum_{i\in S(n)}\frac{1}{i}}{\sum_{i\in \psi_jS(n)}\frac{1}{i}} = 1$ for every $j$, $n\in\mathbb{N}$. Return to main problem. Consider any $a\in A$ we have that $S(a)\subseteq A$ and for every $i\not= j$, $\psi_iS(a)\cap\psi_jS(a)=0$. From lemmas 1, 2 we see that $1=\lim_{a\in A, a\to\infty}\frac{\sum_{k=1}^n\sum_{i\in \psi_kS(a)}\frac{1}{i}}{\sum_{k=1}^n\frac{1}{b_k}\sum_{i\in S(a)}\frac{1}{i}} \leq \lim_{a\in A, a\to\infty}\frac{\sum_{i\in S(a)}\frac{1}{i}}{\sum_{k=1}^n\frac{1}{b_k}\sum_{i\in S(a)}\frac{1}{i}} = \frac{1}{\sum_{k=1}^n\frac{1}{b_k}}$, so $\sum_{k=1}^n\frac{1}{b_k}\leq 1$. $\Box$ Moreover from this proof we see that if $A=\cup b_iA + c_i$, then $\frac{1}{b_1}+\ldots +\frac{1}{b_n} = 1$.

Is this correct? Limits always go haywire...

Hmmm. Is that a calculus solution ?

Solution with stroller Consider the directed graph with vertices the elements of $A$, and edge $u\to v$ iff $v=b_iu+c_i$ for some $i$. The condition implies that every vertex has outdegree $n$, and indegree $1$, so we get a forest, with each connected component generated by its root.* Note that we only need to prove the result on one connected component, so suppose $A$ corresponds to the minimal set generated by root $a$. Now, suppose we have a vertex $v$ such that $a\to v$ has $e_i$ $i$-edges. Then, $a\le C\prod b_i^{e_i}$ for some constant $C$. So, if we consider all vertices with values $\le C*n^k$, this corresponds to at least the set of vertices such that $\sum e_i\log_n{b_i}\le k$. Call this set $A_k$. If we increment $k$, the $i$-edge of a given chain will increment by $\log_{b_i}n$ on average. So, each vertex will yield $\sum \log_{b_i}n$ children every increment, on average, and hence $|A_k|$ is of the order $\left(\sum\log_{b_i}n\right)^k$. However, by Jenson's on $\frac{1}{\ln(1/x)}$, $$\sum\frac{1}{b_i}>1\implies \sum\log_{b_i} n=\ln n\sum\frac{1}{\ln b_i}>\ln(n)\cdot\frac{n}{\ln n}=n$$So, $|A_k|$ grows faster than $n^k$. However, we know that all elements in $|A_k|$ are less than $Cn^k$, so we eventually have repeated elements in $A_k$, which is a contradiction, as desired. *See below for diagram

\(\mathcal{YUH}\).

The following solution is a bit hastily written; which I apologize for. Also I am typing it late at night, so please do highlight if you find any errors in the solution.