<s>[INST] Summing ${\frac{1}{n^2}}$ over subsets of $N$. Are there $2$ subsets, say, $A$ and $B$, of the naturals such that
$$\sum_{n\in A}   f(n)   =  \sum_{n\in B}   f(n)$$
where $f(n)={\frac{1}{n^2}}$?
If $f(n)=\frac{1}{n}$ then there are many counterexamples, which is probably a consequence of the fact that the harmonic series diverges:
$$\frac23 = \frac12 + \frac16 = \frac14+\frac13+\frac1{12}$$
And if $f(n)=b^{-n}$ for some base $b$ then it is true because for all $M$, $\sum_{n>M} f(n) < f(M)$. (This is just the base-b representation of a real number.The case $b=2$ gives a bijection surjection $2^{N} \to [0,1])$.
So we have sort of an in-between case here.
Also, what if $A$,$B$:
-are required to be finite sets?
-are required to be infinite and disjoint?
 [/INST] Yes to both cases:
1) $\frac{1}{15^2}+\frac{1}{20^2}=\frac{1}{12^2}$
2) $\frac{1}{15^2}+\frac{1}{150^2}+\frac{1}{1500^2}+...+\frac{1}{20^2}+\frac{1}{200^2}+\frac{1}{2000^2}+...=\frac{1}{12^2}+\frac{1}{120^2}+\frac{1}{1200^2}...$
for first case - if we have pythagorean triple (a,b,c), such that $a^2+b^2=c^2$, then:
$\frac{1}{a^2 b^2}=\frac{1}{a^2 c^2}+\frac{1}{b^2 c^2}$
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