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About LinkBacksProve that:
^{r^i} = \lim_{n\rightarrow \infty} \prod_{j = 0}^{1}\sum_{i = 0}^{n} (a^{j}\cdot r)^i)
to reverse summation order we only play with inqualities. (this trick is also useful when reversing integration order on double integrals.)
thus, for both sums, we have&
now just combine these inequalities to get
and finally split this multiple inequality into two ones to get the desired order of summation.
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