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 $B=A+A=\{x+y\mid x\in A, y\in A\}$.
Construct $A\subset\mathbb N$ with $f(A)=\sum\limits_{x\in A}\frac1x$ finite and $f(A+A)$ infinite
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0What did you try? There is a simple, very tempting, case... – 2012-07-15
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0I can't prove the divergence of $f(B)$... – 2012-07-15
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0This is not my question: please explain what case(s) you tried. – 2012-07-15
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0$1/2^n$ or $1/n^2$ – 2012-07-15
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0I'm not sure whether these two cases satisfy the requirement of the problem – 2012-07-15
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0I 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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0emm..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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0Are 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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0Golbez: Well done. – 2012-08-17
2 Answers
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.
- 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 $p\equiv1\pmod{4}$.
- 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 $p\equiv a\pmod{b}$.
- For $a=1$, $b=4$, this shows that $f(A+A)$ diverges.
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0That's nice :-) – 2012-07-15
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0@joriki: $\langle$ *Blushes* $\rangle$. – 2012-07-15
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0Good 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 holds – 2012-07-15
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0You 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
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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0Nice solution. Thanks for introducing Kempner series. – 2016-09-15