Rare Gems

Let the proportion of the $1^\text{st}$ gemstone be $k$ , then
the proportion of the $2^\text{nd}$ gemstone is $\dfrac12 \times k = \dfrac k{2!}$ ,
the proportion of the $3^\text{rd}$ gemstone is $\dfrac13 \times \dfrac k{2!} = \dfrac k{3!}$ ,
the proportion of the $4^\text{th}$ gemstone is $\dfrac14 \times \dfrac k{3!} = \dfrac k{4!}$ ,
and so on.

This suggests that the proportion of the $n^\text{th}$ gemstone is $\dfrac k{n!}$ , where $n$ is a positive integer.

Let us prove the claim above by induction :

Basic step : $n=1$ is true because the proportion of the $1^\text{st}$ gemstone is $k = \dfrac k{1!}$ .

Inductive step : Assume $n=m$ is true for some positive integer $m$ , then the proportion of the $m^\text{th}$ gemstone must be $\dfrac k{m!}$ . Extending this shows that the $(m+1)^\text{th}$ gemstone must be $\dfrac1{m+1} \times \dfrac k{m!} = \dfrac k{(m+1)!}$ . So the proposition for $n=m+1$ is true as well. Thus our claim is indeed true.

Now let's calculate the proportion of the first gemstone (as compared to the entire population of gemstones):

$\begin{aligned} \dfrac{\text{ proportion of the first gemstnone } }{ \text{ total proportion of all the gemstones } } &=&\dfrac k { \dfrac k{1!} + \dfrac k{2!} + \dfrac k {3!} + \dfrac k {4!} + \cdots } \\ &= & \dfrac {\cancel k} { \dfrac {\cancel k} {1!} + \dfrac {\cancel k} {2!} + \dfrac {\cancel k} {3!} + \dfrac {\cancel k} {4!} + \cdots } \\ &= & \dfrac1 { \dfrac 1{1!} + \dfrac 1 {2!} + \dfrac 1 {3!} + \dfrac 1 {4!} + \cdots } \\ &= & \dfrac1 { -\dfrac1{0!} + \left( \dfrac1{0!} + \dfrac 1{1!} + \dfrac 1 {2!} + \dfrac 1 {3!} + \dfrac 1 {4!} + \cdots \right) } \\ &= & \dfrac1 { -\dfrac1{0!} + e^1 } = \dfrac1{e-1} \approx \boxed{0.582} \\ \end{aligned}$

Note that the series of $e^x$ can be expressed as $\displaystyle \sum_{r=0}^\infty \dfrac{x^r}{r!}$ , so $e^1 = \dfrac1{0!} + \dfrac 1{1!} + \dfrac 1 {2!} + \dfrac 1 {3!} + \dfrac 1 {4!} + \cdots$ .