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We know that sum of natural numbers over any interval always exists.
For example sum from 0 to 10 of all natural numbers is $$S=\sum_{n=0}^{10}{n}=\frac{0+10}{2}\times{10}=55$$
But what about real numbers?
Is sum of all real numbers from 0 to 10 infinite?
If it is infinite how to prove it?
$$ \sum_{n \atop {\vphantom{\Huge A}x_{n}\ \in\ \left[0,1\right]}}x_{n} > \sum_{n = 1}^{\infty}{1 \over n} $$
How would you sum up all the real numbers between $0$ and $10$? You'd have to assemble all the real numbers in $[0,10]$ and then add up their values. But there is an infinite number of reals just in $[0,1]$, so you can't assemble all of them, by definition. So there is no way for you to sum all the reals between $0$ and $10$.
That's an intuitive explanation. We can deliver an actual proof:
Suppose, for contradiction, that the sum $n$ of all reals in $[0,10]$ is defined. This sum implies that we have a finite series converging to $n$, because the starting and ending terms of the interval are defined. A finite series is the sum of a sequence that has a finite number of terms. Take our sequence $(a_k)$ with its finite number of terms. As the sequence is finite, let the terms be ordered such that the sequence is monotonically strictly increasing. (Obviously we will not have two equal terms in the sequence.) Take any two adjacent terms $a_x, a_{x+1}$. The rationals are dense everywhere in the reals, so between $a_x,a_{x+1}$ there exists some $\frac pq$ s.t. $a_x < \frac pq < a_{x+1}$. As $\frac pq$ is a rational, it is also a real, so there exists a real that is not in the sequence, so the sequence does not sum all the reals in the interval.
