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While playing with expressions, I came up to the following "infinite sums". I haven't seen them anywhere else but maybe I didn't look long enough. Find the values of $s$ and $t$. $$s=\sum\nolimits_{1}^{\sum\nolimits_{1}^{\sum\nolimits_{1}^{~\mathstrut^{.^{.^{.^{.^{}}}}}~}1}1}1$$ $$t=\int_{0}^{\int_{0}^{\int_{0}^{~\mathstrut^{.^{.^{.^{.^{}}}}}~} dx} dx} dx\\$$ Does this mean that notation is inadequate at representing values in a nonambiguous way or are these expressions just an example of notation abuse?

Servaes
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Tos
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  • Related: https://math.stackexchange.com/questions/1623403/references-about-iterating-integration-int-a-0-int-a-1-vdots-i-1dxi-0 – Watson Jun 07 '16 at 11:44

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It is not entirely unambiguously clear to me what those expressions should mean. The dots suggest something of a limiting process, so I'd interpret them as limits.

The expression for $s$ can be viewed as a limit as follows: Define a sequence $s_0,s_1,\ldots$ as $$s_0:=1\qquad\text{ and }\qquad s_{n+1}:=\sum_{i=1}^{s_n}1\ \text{ for }\ n\geq0.$$ Then $s$ can be viewed as the limit of this sequence, i.e. $s=\lim_{n\to\infty}s_n$. But the expression for $s_{n+1}$ can be written as $$s_{n+1}=\sum_{i=1}^{s_n}1=s_n,$$ so we have $s_n=s_0=1$ for all $n$, and hence $s=\lim_{n\to\infty}=1$. Perhaps the choice of $s_0:=1$ was a bit arbitrary, but for any choice of $s_0$ we get $s=s_0$ by the same reasoning.

For $t$ an entirely analogous argument shows an entirely analogous result.

So I would say that these expressions, without further explanation, are notational abuse. It is not unambiguously clear what the dots should mean, and if there is a limit involved in the way I described here, it is not clear what the initial value should be. Moreover, the limit is equal to the initial value in these particular cases, so the notation also seems useless, apart from making a fun exercise.

Servaes
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Without thinking too much of how they would be defined, you are essentially looking for solutions to $$ s = \sum_{n=1}^s 1 $$ and $$ t = \int_0^t dx $$ which both can be any value at all (although they must be integral in the first case). So they are certainly not unique...

Simon Rose
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