Sum of reciprocals of odd numbers that add up to 1 I am trying to find an Egyptian factions expansion of 1: all the odd denominators are distinct ones; $o_k<o_{k+1}$; the last or maximum odd number in the denominator is the product of all other odd denominators.
$$1=\dfrac{1}{o_1}+\dfrac{1}{o_2}+\cdots+\dfrac{1}{o_{n}}+\dfrac{1}{\prod\limits_{{\rm{k}} = 1}^n {{o_k}} }$$
Does this expansion exist? if not, can it be proved? if it exists, can there be any example of such an expansion?
If it allows that $n\to\infty$, then can there exist such an infinite series version expansion?
 A: ${1\over3}+{1\over9}+{1\over27}+\cdots={1\over2}$;
${1\over5}+{1\over25}+{1\over125}+\cdots={1\over4}$;
${1\over7}+{1\over49}+{1\over343}+\cdots={1\over6}$;
${1\over13}+{1\over169}+{1\over2197}+\cdots={1\over12}$;
${1\over2}+{1\over4}+{1\over6}+{1\over12}=1$,
so that's how you can do it with an infinite series of reciprocals of distinct odd integers. 
For the finite sum, there's an example at Can the sum of finitely many inverses of distinct odd integers $\geq 3$ be 1? but it doesn't satisfy the last term being the product of all the others. 
A: Further to the example in the comments, where $o_1=3$ and $o_2=5$, leading to the sum being $\frac 35$, we can generalise the counter example:
Lets work firstly with $2$ numbers, $o_1=2k+1$ and $o_2=2m+1$, where $k<m$. We therefore have the sum
\begin{align}S&=\frac1{2k+1}+\frac1{2m+1}+\frac1{(2k+1)(2m+1)}\\\\
&= \frac{2m+1}{(2k+1)(2m+1)}+\frac{2k+1}{(2m+1)(2k+1)}+\frac1{(2k+1)(2m+1)}\\\\
&=\frac{2m+1+2k+1+1}{(2k+1)(2m+1)}\\\\
&=\frac{2m+2k+3}{4km+2k+2m+1}\end{align}
For this to be equal to $1$ then we must have \begin{align}2m+2k+3 &= 4km+2k+2m+1\\
3&=4km+1\\
4km &= 2\\
km &= \frac 12\end{align}
Give you have specified that $o_1$ and $o_2$ have to be odd numbers, then $k$ and $m$ must be whole numbers and we cannot satisfy the above equation, $km=\frac 12$ and the sum can never be equal to $1$.

In the case that we have $o_i$ for $i=1$ to $n$, we can say that each $o_i$ is equal to $2k_i+1$ where $k_i < k_i+1$
We can write the sum as \begin{align}S&=\frac1{2k_1+1} + \frac1{2k_2+1} + \cdots + \frac 1{2k_n+1} + \frac 1{\Pi_{i=1}^n(2k_i+1)}\\\\
&=\frac{\Pi_{i\neq 1}(2k_i+1)}{\Pi_{i=1}^n(2k_i+1)} + \frac{\Pi_{i\neq 2}(2k_i+1)}{\Pi_{i=1}^n(2k_i+1)}+\cdots+\frac{\Pi_{i\neq n}(2k_i+1)}{\Pi_{i=1}^n(2k_i+1)} + \frac 1{\Pi_{i=1}^n(2k_i+1)}\\\\
&= \frac{\Pi_{i\neq 1}(2k_i+1)+\Pi_{i\neq 2}(2k_i+1)+\cdots +\Pi_{i\neq n}(2k_i+1)+1}{\Pi_{i=1}^n(2k_i+1)}\end{align}
The proof here that the numerator and denominator cannot be equal is quite tricky so I will leave it to you for now and attempt to add a proof when I have the time later
A: This is a special case of improper Znám’s problem which is still unsolved according to the Wiki ：https://en.wikipedia.org/wiki/Zn%C3%A1m%27s_problem
Check http://primefan.tripod.com/ZnamProbSols.html for solutions of Znám’s problem.All known ones contain one even number,so don't fit the condition that all numbers are odd.
However,If we replace the product in the denominator of last term by least common multiple，Then there are infinity many solutions.
{3, 5, 7, 9, 11, 15, 21, 135, 10395} is one.
Set last number $o_N=l.c.m(o_1,\dots,o_{N-1})=ab,1<a<b$,Since
$$\frac{1}{ab}=\frac{1}{a(1+a+b)}+\frac{1}{b(1+a+b)}+\frac{1}{ab(1+a+b)},$$and $l.c.m(ab,a(1+a+b),b(1+a+b))=ab(1+a+b)$,we can span the last term into 3 terms.Keep doing this you'll get the infinity many solutions.
Reference:
R.Arce-Nazario, F.Castro, R.Figueroa. the number of solutions of $\sum_{i=1}^{11}\frac{1}{x_i}=1$ in distinct odd
natural numbers. http://dx.doi.org/10.1016/j.jnt.2012.11.011
