We roll a fair die until a $5$ appears. What is the expected value of the minimum value rolled? 
Question: Given a fair dice, we roll until we get a $5.$ What is the expected value of the minimum value rolled?

Answer is $\frac{137}{60}.$
There is a similar question asked in MSE but I do not understand the method used by Henry. 
In particular, if we let $X$ be the minimum value rolled up to and including $5$, then 
$$E(X) = \sum_{x=1}^5 xP(X=x) = 1 \times \frac12 + 2 \times \frac16 + 3 \times \frac1{12}+4 \times \frac{1}{20}+5 \times \frac15 = \frac{137}{60}.$$
It seems that we are using the fact that 
$$P(X=x) = \frac{1}{x(x+1)}.$$

I do not understand how to obtain the equation above.


Assume that a $5$ is first seen on roll $n$.
$5$ is the lowest seen with $n-1$ $6$s then one $5$.
$4$ is the lowest seen with $n-1$ $4$s and $6$s, but not all $6$s then one $5$.
$3$ is the lowest seen with $n-1$ $3$s, $4$s, and $6$s, but not all $4$s and $6$s then one $5$.
$2$ is the lowest seen with $n-1$ $2$s, $3$s, $4$s, and $6$s, but not all $3$s, $4$s, and $6$s then one $5$.
$1$ is the lowest seen with $n-1$ $1$s, $2$s, $3$s, $4$s, and $6$s, but not all $2$s, $3$s, $4$s, and $6$s then one $5$.
$$
\begin{array}{c|l|l}
\text{lowest}&\text{chance with $n$ rolls}&\text{sum over $n$}\\
\hline
5&\,\left(\frac16\right)^{n-1}\frac16&\frac15\\
4&\,\left[\left(\frac26\right)^{n-1}-\left(\frac16\right)^{n-1}\right]\frac16&\frac1{20}\\
3&\,\left[\left(\frac36\right)^{n-1}-\left(\frac26\right)^{n-1}\right]\frac16&\frac1{12}\\
2&\,\left[\left(\frac46\right)^{n-1}-\left(\frac36\right)^{n-1}\right]\frac16&\frac16\\
1&\,\left[\left(\frac56\right)^{n-1}-\left(\frac46\right)^{n-1}\right]\frac16&\frac12
\end{array}
$$
Expected value $=5\cdot\frac15+4\cdot\frac1{20}+3\cdot\frac1{12}+2\cdot\frac16+1\cdot\frac12=\frac{137}{60}$
