# Summing ${\frac{1}{n^2}}$ over parts of $N$.

Exist $2$ parts, claim, $A$ and also $B$, 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 $f(n)=\frac{1}{n}$ after that there are several counterexamples, which is possibly an effect of the reality that the harmonic collection deviates : $$\frac23 = \frac12 + \frac16 = \frac14+\frac13+\frac1{12}$$
And if $f(n)=b^{-n}$ for some base $b$ after that it holds true due to the fact that for all $M$, $\sum_{n>M} f(n) < f(M)$. (This is simply the base-b depiction of an actual number.The instance $b=2$ offers a bijection surjection $2^{N} \to [0,1])$.

So we have type of an in-between instance below.
Additionally, what happens if $A$, $B$ :
- are called for to be limited collections?
- are called for to be boundless and also disjoint?

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2019-05-07 13:09:40
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Yes to both instances : 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 instance - if we have pythagorean three-way (a, b, c), such that $a^2+b^2=c^2$, after that : $\frac{1}{a^2 b^2}=\frac{1}{a^2 c^2}+\frac{1}{b^2 c^2}$