# Summing infinitesimals to 1

Tl;dr: Could someone give me an explanation as to how infinitesimal numbers might be added together to get 1?

Why I want to know: some people (philosophers) like to introduce infinitesimals into probability theory when considering cases like an infinite amount of coin tosses. To say that the event that this coin will land heads every time has probability 0, they think, is bad, as probability 0 should be reserved for the impossible. Instead, they say it should have an infinitesimal probability.

Fair enough. But presumably, the probability that the coin corresponds to some infinite sequence of heads and tails is 1. If we like countable additivity, this means the probability of all the possible infinite sequences should sum up to 1.

Apparently this is consistent with their theory. Each infinitesimal probability of each infinite sequence will, when added together, equal 1.

I'm confused as as to how this could happen. Why would an infinite amount of infinitesimals add up to 1, and not, say, 2?

• I'd say they are unreasonable, but that's just me. After all you are summing uncountably many "infinitesimals", which does not make sense to begin with. – Klaus Oct 9 at 18:21
• @Klaus $\int_0^1\,dx = 1$ – amsmath Oct 9 at 18:22
• "Apparently this is consistent with their theory." I'd be interested to see it, because this doesn't pass the smell test for me. It sounds much more like physics-speak, where things are not quite rigorous but rigor-ish and can break down when pressed into corner cases. One thing to be careful of here: you talk about infinite sequences of coin flips and using countable additivity on their probabilities. But note that there are in fact an uncountable number of infinite strings of coin flips, so countable additivity would seem not to be of much help. Calculus, on the other hand, would. – Aaron Montgomery Oct 9 at 18:31
• @amsmath IMO that's a formalism that avoids the question. How does adding together infinitely many infinitesimal things actually work? – user4894 Oct 9 at 18:32
• At least for Laplace-experiments, it doesn't work out in any setting of infinitesimals I know. What however works is that instead of infinite sequences, we look at finite sequences in the new setting, which are longer than any finite sequence in a standard setting. Depending on what you want to achieve however, this approximation can be useful or useless. – Sudix Oct 9 at 18:44

Note that the geometric series $$1/2 + 1/4 +1/8+.... =\sum _1 ^\infty \frac {1}{2^n} =1$$
More interesting is for a small positive real number $$\epsilon$$ $$\sum _1 ^\infty \frac {\epsilon}{2^n} =\epsilon$$
The short answer is you need to integrate up the infinitesimals, rather than summing them. This is because infinitesimal probabilities emerge from continuous distributions. With a pdf $$f(x)$$, $$f(x)dx$$ is an infinitesimal probability for $$x$$ existing in a width-$$dx$$ range of values.
We can flesh this out. Take an infinite sequence of coin tosses and label the tosses with integers $$\ge1$$. If $$S$$ is the set of tosses yielding heads, associate the sequence with $$x(S):=\sum_{n\in S}2^{-n}\in[0,\,1]$$. In countably infinitely many cases, a number is achievable with multiple sequences; but integration doesn't care about that. For any subset $$T$$ of $$[0,\,1]$$ for which $$\int_Tdx$$ is defined, we can say that integral is the probability of the sequence's representation being an element of $$T$$. Unitarity is then the condition $$\int_0^1dx(S)=1$$.