we begin with the numbers $1,\frac{1}2 ,\frac{1}3,\ldots \frac{1}{100}$ written in a board. We do the following operation : we delete $2$ numbers $a$ and $b$ from the board , and we remplace them with the number $a+b+ab$. what is the maximal value that we can have in the board after $99$ operations
here is what i think if we have for example numbers $a$ and $b$ and $c$ in the board if we choose $a$ and $b$ first we will have two numbers , $(a+ab+b,c)$ then we will have the number $a+b+c+ab+bc+ac+abc$ and if we choose $a$ and $c$ first we will have $a+c+ac$ and $b$ , then $a+b+c+ab+bc+ac+abc$ ...... i think that whatever numbers we choose we will have the same number at the end , but we need to prove it , and we need to know the number that will remain.