<s>[INST] Generalized version of calculating expected minimum value rolled until $5$ is obtained from a fair die The following is a modified interview question.

Given an $n$-sided fair die where $n\geq 1.$ 
  You roll a die until you get a $m$ where $1\leq m\leq n$. 
  Calculate the expected value of the minimum rolled. 

The original interview question takes $n=6$ (standard fair die) and $m=5$. 
I mange to solve the problem and I reproduce my attempt below.
The expected minimum value rolled is $\frac{137}{60}$ because if $X$ is the minimum value rolled up to and including $5,$ then 
$$P(X=x) = \frac{1}{x(x+1)} \quad \text{for }x=1,2,3,4 \quad \text{and} \quad P(X=5) = \frac{1}{5}.$$
So, 
$$E(X) = \sum_{x=1}^5 xP(X=x) = \frac{137}{60}.$$
I am trying to solve the generalized version of the problem.

By the same spirit, let $Y$ be the minimum value rolled up to and including $m.$ Then 
$$P(Y=y) = \frac{(y-1)!}{(y+1)!} = \frac{1}{y(y+1)} \quad \text{for }y =1,2,...,m-1 \quad \text{and}\quad P(Y=m) = \frac{1}{m}.$$
Therefore,
$$E(Y) = \sum_{y=1}^m y P(Y=y) = 1 + \sum_{y=1}^{m-1} \frac{1}{y+1} = 1 + \sum_{y=2}^{m} \frac{1}{y}.$$
 [/INST] Your work looks correct.
A slightly different approach: for $1 \le y \le m$,
$$P(Y \ge y) = P(\text{all rolls are $\ge y$}) =\sum_{k=0}^\infty \left(\frac{n-y}{n}\right)^k \frac{1}{n} = \frac{1}{y}.$$
Then since $Y$ is nonnegative,
$$E[Y] = \sum_{y=1}^m P(Y \ge y) = \sum_{y=1}^m \frac{1}{y}.$$
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