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For every $U\subset\mathbb N$, define $$f(U)=\sum_{x \in U} \frac{1}{x}$$ Construct $A$ such that $f(A)$ is finite and $f(B)$ is infinite, with $B=A+A=\{x+y\mid x\in A, y\in A\}$.

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    What did you try? There is a simple, very tempting, case...2012-07-15
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    I can't prove the divergence of $f(B)$...2012-07-15
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    This is not my question: please explain what case(s) you tried.2012-07-15
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    $1/2^n$ or $1/n^2$2012-07-15
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    I'm not sure whether these two cases satisfy the requirement of the problem2012-07-15
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    I assume you mean (1.) $A=\{2^n\mid n\in\mathbb N\}$ and (2.) $A=\{n^2\mid n\in\mathbb N\}$. Obviously, $f(A)$ is finite in both cases (good). But case (1.) leads to a set $B$ which is too *thin* for $f(B)$ to diverge, so let us forget it. What did you try in case (2.)?2012-07-15
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    emm..I tried to find an inequality. let $a_{n}=\sum_{i=1}^{+\infty} \frac{1}{n^2+i^2}$ ,and show that $S_n$ is divergent, which is equivalent to the divergence of $f(B)$.2012-07-15
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    Are those equivalent? You're only summing over the *distinct* values in $B$, so be careful not to count anything twice...2012-07-15
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    Golbez: Well done.2012-08-17

2 Answers 2

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Let $A=\{n^2\mid n\in\mathbb N\}$, then $f(A)$ is obviously finite. Let us show that $f(A+A)$ is infinite. Note that $f(A+A)$ is not in general $\sum\limits_{n\in A}\sum\limits_{k\in A}\frac1{n+k}$ since the double sum might (and often does) count several times many elements of $A+A$. Thus, we must exhibit sufficiently many elements in $A+A$ and count each of them only once. Here is a way to do this.

  1. A famous theorem due to Fermat asserts that every prime $p$ such that $p\equiv1\pmod{4}$ is a sum of two squares (and that the other primes are not). Thus, $f(A+A)\geqslant\sum\limits_p\frac1p$, where the series enumerates the primes $p$ such that $p\equiv1\pmod{4}$.
  2. An equally famous theorem due to Dirichlet asserts that, for every $a$ and $b$ with no common factor, $\sum\limits_p\frac1p$ diverges, where the series enumerates the primes $p$ such that $p\equiv a\pmod{b}$.
  3. For $a=1$, $b=4$, this shows that $f(A+A)$ diverges.
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    That's nice :-)2012-07-15
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    @joriki: $\langle$ *Blushes* $\rangle$.2012-07-15
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    Good job! I found another way to do that, Let $a_n=\sum_{i=1}^{+\infty} \frac{1}{n^2+i^2}$, it's easy to know that $S_n $ converges when $f(B)$ converges. When $\sum_{n=1}^{+\infty} \sum_{i=1}^{+\infty} \frac{1}{n^2+i^2}>\sum_{n=1}^{+\infty} \sum_{i=1}^{+\infty} \frac{1}{(n+i)^2}>\sum_{n=1}^{+\infty} \frac{n}{4n^2}>+\infty$ also holds2012-07-15
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    You seem to have purely and simply skipped the warning at the beginning of my post (also apparent in @Ben Millwood's comment), which, unfortunately, destroys your "argument". (Unrelated: you do not define $S_n$.)2012-07-15
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Let $A$ be the set of positive integers whose decimal expansion is missing a certain digit, say the digit $9$. It is a standard but interesting fact that the sum of the reciprocals of the elements of $A$ converges. The proof uses little machinery.

The set $A+A$ consists of the integers $\ge 2$, so the sum of the reciprocals of elements of $A+A$ diverges.

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    Nice solution. Thanks for introducing Kempner series.2016-09-15