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I want to rigorously prove that given any $x\geq 0$, we can write $x=\sum_{k\in J} \frac{1}{k}$ for some $J\subseteq \mathbb{N}$ eg. $2=\sum \frac{1}{2^{k}}$.

Also, for what other $\{a_{k}\geq 0\}$ is that true (any theorems?)? For example, it fails for $a_{k}=\frac{1}{k!}$: Writing number as sum of reciprocals of factorial

Another possible $\{a_{k}\}$ collection is $\mathbb{N}\cup \{\frac{1}{2^{n}}\}$.

For conditionally convergent we have the Riemann series theorem but in our case $\sum a_{k}$ is not even conditionally convergent.

Anyhow back to proving it for reciprocals.

So there is a bunch of algorithms https://en.wikipedia.org/wiki/Greedy_algorithm_for_Egyptian_fractions

and a proof of what I want in "The Approximation of Numbers as Sums of Reciprocals". But the proof there is a bit too wordy for me. I bet if I sit down I can flesh it out completely.

Attempt

1)First for rationals $\frac{x}{y}$, we split into $\frac{x}{y}=\frac{1}{\lceil y/x\rceil}+\frac{(-y)\bmod x}{y\lceil y/x\rceil}$, then repeat for $\frac{(-y)\bmod x}{y\lceil y/x\rceil}$ and the result follows by induction on x ($x=1$ and then $x>1$):

$\frac{x+1}{y}-\frac{1}{\lceil y/x+1\rceil}=\frac{(x+1)\lceil y/x+1\rceil-y}{y\lceil y/x+1\rceil}\leq \frac{(x+1)( \frac{y-1}{x+1}+1)-y}{y\lceil y/x+1\rceil}=\frac{x}{y\lceil y/x+1\rceil}$.

Thus, $(x+1)\lceil y/x+1\rceil-y<x$, for which the result is already true by IH.

Then because any irrational (eg. decimal expansion) can be written as sum of rationals we are done. A possible problem here is repetition of reciprocals.

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  • $\begingroup$ If $J \subset \mathbb{N}$ then you cannot write any irrational number in the sum of $\frac{1}{k}$ because sum of rational numbers will give you a rational number. $\endgroup$
    – gamma
    Sep 28 '15 at 17:05
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    $\begingroup$ J can be infinite eg. $e=0!+\frac{1}{1!}+\frac{1}{2!}+...$. $\endgroup$
    – user133100
    Sep 28 '15 at 17:07
  • $\begingroup$ How does your proof for rationals guarantee distinct denominators? $\endgroup$
    – Aravind
    Sep 28 '15 at 18:44
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If $0<x<1$, then the reciprocals of the powers of 2 suffice. Otherwise, find $N$ such that $0<|x-\sum_{i=1}^{N}\frac{1}{(2i-1)}|<1$. So $J$ is the union of odd numbers and powers of 2.

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  • $\begingroup$ what is that N explicitly or why does it exist? $\endgroup$
    – user133100
    Sep 28 '15 at 19:01
  • $\begingroup$ It exists because the sum of reciprocals of the odd numbers diverges. $\endgroup$
    – Aravind
    Sep 28 '15 at 19:02
  • $\begingroup$ Yeah, but why does it follow from that? Sum of reciprocal factorials go to infinity too. $\endgroup$
    – user133100
    Sep 28 '15 at 19:03
  • $\begingroup$ There's a $M$ such that $\sum_{i=1}^{M}\frac{1}{2i-1}>x$; hence there's a minimum such $M$; so let $N=M-1$. $\endgroup$
    – Aravind
    Sep 28 '15 at 19:14
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I think this may work:

Let $x>0.$ Let $n_0$ be the largest integer $t$ for which $t\leqslant x.$ Having chosen $n_0,\ldots,n_m,$ let $n_{m+1}$ be the largest integer $t$ for which $\dfrac{n_0}{10^0}+\cdots+\dfrac{n_{m+1}}{10^{m+1}}\leqslant x.$ Then $x=\sum\limits_{m=0}^\infty\dfrac{n_m}{10^m}.$ Since every rational number is an egyptian fraction, then each $\dfrac{n_m}{10^m}$ is an egyptian fraction and the result follows.

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  • $\begingroup$ They might be repetitions of egyptian fractions and so you might get integers. $\endgroup$
    – user133100
    Sep 28 '15 at 19:49
  • $\begingroup$ Then we express one of the repeated numbers as an egyptian fraction and so on until we have something "nice" (I'm being too "chatty" here of course) $\endgroup$
    – CIJ
    Sep 28 '15 at 19:56

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