Expected value of Distinct Elements from maximum sequence of permutation of 1 to n

Let $$a_1,a_2 ,... a_n$$ be a permutation of 1 to $$n$$. We define sequence $$b = b_1 , b_2 ,... ,b_n$$ as $$b_i = max ~~{a_1,a_2,...a_i}$$. Find Expected Value of $$X$$: distinct numbers in $$b_i$$.

For example if the permutation is $$1, 3, 2$$ then $$b = 1,3,3$$ so X is 2. (i.e 1 and 3 are distinct numbers). Another example: for permutation $$1 , 3 , 4 ,2$$ we have $$b = 1,3,4,4$$ so $$X = 3$$.

If we do this for all permutations of 1 to 3, we get sum of distinct numbers in $$b$$: $$11$$. For permutation of $$1$$ to $$4$$, it is $$50$$. So expected values are $$11/6$$ and $$50/24$$ respectively.

But I cannot find the pattern in this to find the answer for general case $$n$$. I tried to use an indicator variable $$I_i = \begin{cases} 1, & \text{if i appears in position 1 to i} \\[2ex] 0, & \text{o.w} \end{cases}$$

And tried to define $$X_ = \sum I_i$$ but I found out this is not correct.

So I want help for this question. This problem is probably related to linearity of expectation.

Let $$I_i$$ be the indicator variable that $$a_i = b_i$$, namely that $$a_i$$ is the maximum of the first $$i$$ values of the permutation.
Hint: Show that $$X = \sum I_i$$.
Hint: Show that $$E[I_i] = \frac{1}{i}$$.
Hence, $$E[X] = \sum_{i=1}^n \frac{1}{i}$$.
This agrees with the $$n=3, 4$$ cases that you calculated.