If a multiple-choice test (where all questions are worth the same) has $n$ choices on each question, random guessing gets you a score of $\frac1n$ on average. But what if the number of choices increased as you went through the test? Let's say that the number of choices matches the question number, so question $\#3$ has $3$ choices, and question $\#5$ has $5$ choices, and so on. If this test has $n$ questions, what is the expected score for random guessing?
The way I think of it is letting each question be worth 1 point and $X_i$ represent the amount of points gained on question $\#i$. Then the expected value for $X_i$ is $$E(X_i)=0\cdot\frac{i-1}{i}+1\cdot\frac1i=\frac1i$$
So the expected value for the amount of points gained on the entire test is $$E(X)=E(X_1+X_2+\dots+X_n)$$ $$E(X)=E(X_1)+E(X_2)+\dots+E(X_n)$$ $$E(X)=1+\frac12+\dots+\frac1n=H_n$$
Since there are n points in total, the expected score is $H_n/n$.
Is this the correct way to find the expected score on the test? If it is, is there an approximation or limit for $H_n/n$?
Yes, this is correct. $H_n\sim \ln n$ as $n\to\infty$, that is, the ratio $H_n/\ln n\to 1$. One way to see this is that $\ln n=\int_1^n\frac1xdx$, and by approximating the area by $n$ rectangles you can show that $1+\frac12+\cdots+\frac1n>\ln n>\frac12+\frac13+\cdots \frac1{n+1}$, so $\ln n<H_n<\ln n+\frac{n}{n+1}$.
So $H_n/n\approx\frac{\ln n}{n}\to0$.