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,
$$E(Y) = \sum_{y=1}^m y P(Y=y) = 1 + \sum_{y=1}^{m-1} \frac{1}{y+1} = 1 + \sum_{y=2}^{m} \frac{1}{y}.$$
 A: 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,
$$E[Y] = \sum_{y=1}^m P(Y \ge y) = \sum_{y=1}^m \frac{1}{y}.$$
A: A possibly simpler derivation of the fact that $P(Y \geq y) = \frac1y,$
where $Y$ is the smallest number rolled among all the rolls up to and including the first roll of $m.$
Consider the set of numbers $\{1, 2, \ldots, y-1, m\}.$
Let $Z$ be the first number rolled from among the numbers in that set.
(We know at least one number is rolled, since we know that eventually we roll $m$.)
Each number is equally likely to be the first rolled. With probability $\frac1y,$ that number is $m.$
If $m$ is the first number rolled from that set, then $Y \geq y.$
If $m$ is not the first number rolled from that set, then $Y < y.$
Hence $P(Y \geq y) = P(\text{$m$ is rolled before any number less than $y$}) = \frac1y.$
Note that $n$ is irrelevant.
A: A slight variant of angryavian's answer above:
Let $T$ be the number of rolls before the first appearance of $m.$ We have $$P(T = t) = \left(\frac{n-1}{n}\right)^{t}\cdot \frac{1}{n},\ \ t=0,1,2,\ldots.$$
We have $$\mathbb{E}[Y \ | \ T=t] = \sum_{y=0}^{m-1} P(Y> y \ | \ T=t) = \sum_{y=0}^{m-1} \left( \frac{n-y-1}{n-1} \right)^t$$
The last equation comes from the fact that, under the condition $T=t,$ we have $Y>y$ if and only if $X_1, X_2, \ldots, X_t$ are all greater than $y$ (where these $X_i$ may be in $\{1,\ldots, n\} \setminus \{ m\}.$ 
By the Law of Total Expectation, we have
$$\mathbb{E}[Y] = \mathbb{E}\left( \mathbb{E}[Y \ | \ T=t]\right) = \sum_{t=0}^{\infty}  \left( \left(\frac{n-1}{n}\right)^{t}\cdot \frac{1}{n} \ \sum_{y=0}^{m-1} \left( \frac{n-y-1}{n-1} \right)^t \right)$$
$$ = \frac{1}{n} \sum_{y=0}^{m-1} \sum_{t=0}^{\infty} \left(\frac{n-y-1}{n} \right)^t = \frac{1}{n} \sum_{y=0}^{m-1} \frac{1}{1 - \frac{n-y-1}{n}} = \sum_{y=0}^{m-1} \frac{1}{y+1} = H_m$$
