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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 $f(n)=1/n^2$?

If $f(n)=1/n$ 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?

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The reason there are a lot of counterexamples for 1/n has little to do with the fact that the harmonic series diverges; it's just that frequently one can find two fractions of this form whose sum is also of this form. –  Qiaochu Yuan Jul 29 '10 at 8:11
    
Also, binary representation does not give a bijection from 2^N to [0, 1]; the map is surjective but not injective. –  Qiaochu Yuan Jul 29 '10 at 8:17
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2 Answers 2

up vote 9 down vote accepted

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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If two disjoint sets of naturals $A$ and $B$ satisfy $\sum_{n \in A} n^2 = \sum_{n \in B} n^2$, then, if $m = \prod_{n\in A} n \prod_{n \in B} n$, $\sum_{n \in A} (n/m)^2 = \sum_{n \in B} (n/m)^2$ and each of the $n/m$ is the reciprocal of an integer since $n|m$.

This extends directly to higher powers (via copy and paste):

If two disjoint sets of naturals $A$ and $B$ satisfy $\sum_{n \in A} n^k = \sum_{n \in B} n^k$ for a positive integer $k$, then, if $m = \prod_{n\in A} n \prod_{n \in B} n$, $\sum_{n \in A} (n/m)^k = \sum_{n \in B} (n/m)^k$ and each of the $n/m$ is the reciprocal of an integer since $n|m$.

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