Construct $A\subset\mathbb N$ with $f(A)=\sum\limits_{x\in A}\frac1x$ finite and $f(A+A)$ infinite 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 $B=A+A=\{x+y\mid x\in A, y\in A\}$.
 A: 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.


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*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.

A: 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.
