Prove that:

$\displaystyle \lim_{n\rightarrow \infty} \sum_{i = 0}^{n}\left(\sum_{j = 0}^{i} a^j\right)^{r^i} = \lim_{n\rightarrow \infty} \prod_{j = 0}^{1}\sum_{i = 0}^{n} (a^{j}\cdot r)^i$

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- Apr 23rd 2009, 05:43 AMfardeen_genProve that...(limits and series)?
Prove that:

$\displaystyle \lim_{n\rightarrow \infty} \sum_{i = 0}^{n}\left(\sum_{j = 0}^{i} a^j\right)^{r^i} = \lim_{n\rightarrow \infty} \prod_{j = 0}^{1}\sum_{i = 0}^{n} (a^{j}\cdot r)^i$ - Apr 23rd 2009, 06:58 AMKrizalid
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 $\displaystyle 0\le i\le n$ & $\displaystyle 0\le j\le i,$ now just combine these inequalities to get $\displaystyle 0\le j\le i\le n,$ and finally split this multiple inequality into two ones to get the desired order of summation.