Uniform Probability and Riemann Sum

Question

Quoted from the Wikipedia page about Natural Density: We see that this notion can be understood as a kind of probability of choosing a number, which obviously is the reason why Natural Densities are studied in probabilistic number theory.
Standard mathematics says that it is impossible to have a uniform probability distribution on the naturals. Meaning that it is impossible to have a distribution giving equal probabilities to each of (all) the natural numbers.
Not a question, however, about the same for an initial segment of the naturals. The probability for picking, say, the number $7$ out of the initial segment $\{1,2,3,4,5,\dots,n\}$ is simply: $1/n$. The same holds for an arbitrary number $k$ in that segment.
And the sum of all the probabilities is $1/n + 1/n + \dots + 1/n = n.1/n = 1$, as it should.
But something weird happens if we take the limit for $n \rightarrow \infty$. Then each of the probabilities for picking one natural number becomes $$\lim_{n \rightarrow \infty} 1/n = 0$$ while the sum of all probabilities is still $1$, according to $$\lim_{n \rightarrow \infty} 1 = 1$$ How can this happen? A bunch of zeroes that sums up to one? Oh well, it's not just a sum of zeroes. It's an infinite sum of zeroes. But nevertheless.
Let's take another example, that seems only remotely resemblant. The following trivial integral is expressed as the limit of a Riemann sum: $$ \int_0^1 dx = \lim_{n \rightarrow \infty} \sum_{n=1}^n 1/n = \lim_{n \rightarrow \infty} n.1/n = \lim_{n \rightarrow \infty} 1 = 1 $$ Am I blind or what? The only difference with the above "probabilities" is the geometrical interpretation. Here come the probabilities. It is a histogram with height of the blocks $1/n$ and width of the blocks $1$ for $n$ blocks. So the total area of the blocks is $(n.1.1/n) = 1$ :

And here comes the Riemann sum of the trivial integral. The blocks are $1/n$ wide, $1$ high and there are $n$ of them. The total area is $(n.1/n.1) = 1$.

Obviously, $\,\lim_{n \rightarrow \infty} n.1/n\,$ is exactly the same algebraic expression with the integral as with our probabilities. But, on the the contrary, it seems that common mathematics has developed quite different ideas about probabilities. The following is a standard Theorem.
Let $X$ be a random variable which assumes values in a countable infinite set $Q$. We can prove there is no uniform distribution on $Q$ . As follows. Assume there exists such a uniform distribution, that is, there exists $a \ge 0$ such that $P(X = q) = a$ for every $q \in Q$ . Observe that, since $Q$ is countable, by countable additivity of $P$ : $$ 1 = P(X \in Q)=\sum_{q \in Q} P(X = q) = \sum_{q \in Q} a $$ Observe that if $a = 0$ , $\sum_{q \in Q} a = 0$ . Similarly, if $a > 0$ , $\sum_{q \in Q} a = \infty$ . Contradiction.
Let's analyze. The second part of this reasoning is remotely resemblant to the following iterated limit: $$ \lim_{n\rightarrow\infty} n \left[ \lim_{m\rightarrow\infty} \frac{1}{m} \right] = \lim_{n\rightarrow\infty} n\; 0 = 0 $$ First define uniform probabilities. It's easy to see that these will become zero if the initial segment becomes infinite. Then sum up. Evidently the sum then must be zero as well.
Here the variable $m$ has been introduced in order to avoid confusion with the variable $n$.
But maybe this is the cause of the problem. Shouldn't we rather write: $$ \lim_{n\rightarrow\infty} n \left[ \lim_{n\rightarrow\infty} \frac{1}{n} \right] = \lim_{n\rightarrow\infty} n\; 1/n = 1 \qquad \mbox{?} $$ It seems that all of these problems would be solved if it were only possible to define infinitesimal probabilities .. What is right and what is wrong?