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?

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
