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This is maybe a very easy one, but I can't find a solution...

I'm looking for a sequence $a_1,...,a_n$ such that $0\leq a_1<\cdots<a_n<1$ and $\sum_{k=1}^na_k=1$. Of course, this should work for any choice of $n\in\mathbb{N}$.

My first option was a finite geometric series, but couldn't come up with the right parameters.

Another option is to use some discrete probability distribution like Binomial, but this does not guarantee the increasing condition.

Any help will be appreciated.

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  • $\begingroup$ Is ${2/(n^2+n),...,(2n)/(n^2+n)}$ not an answer? $\endgroup$
    – Moonshine
    Commented Jul 8, 2021 at 16:20
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    $\begingroup$ Besides, such $a_1,...,a_n$ won't exist when $n=1$... $\endgroup$
    – Moonshine
    Commented Jul 8, 2021 at 16:20
  • $\begingroup$ Maybe use $1+2+\cdots+n=n(n+1)/2$? $\endgroup$ Commented Jul 8, 2021 at 16:22

2 Answers 2

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Here is a very general, pretty easy way to get what you want.

Take any strictly increasing sequence of $n$ positive numbers. Find their sum. Divide each term by that sum. You now have a strictly increasing sequence of positive numbers wich sums to $1$.

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  • $\begingroup$ Ahh ,normalization. Should have thought of that rather than the Zeno's approach below :). I like!~ $\endgroup$
    – Alan
    Commented Jul 8, 2021 at 16:23
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    $\begingroup$ Sometimes the solution is so easy you just don't see it. Just like searching for the keys that are in your hand. Thanks for the straightforward answer. $\endgroup$
    – RLC
    Commented Jul 8, 2021 at 16:25
  • $\begingroup$ @RLC I often lose my sunglasses on top of my head. Happens to all of us from time to time. $\endgroup$
    – Arthur
    Commented Jul 8, 2021 at 16:26
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Assuming you are okay with a piecewise rule, this is simply done with a modification of the geometric sequence with base $\frac 1 2$. First note that this will be impossible for $n=1$ no matter what sequence you choose. Since you want the sequence ascending, we need to reverse the normal direction of $(\frac 1 2)^n$ which is decreasing. So, for $1\leq i <n$ let $a_i=(\frac 1 2)^{n-i}$ Finally, let $a_n=a_{n-1}+a_{1}$. That gets you that final jump over the Zeno's paradox problem.

Note that this doesn't work for $n=2$ since you won't have inequality, but for $n=2$ you can just use whatever sequence you want like $\frac 1 4, \frac 3 4$

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