Infinite amount of additions, finite sum? I suggest it's a popular question, so if it was asked already (I couldn't find it anyway), close this question instead of downvoting, thanks!
Let's check this addition: $\sum_{n=0}^{\infty}\frac{1}{2^n}=2$
It looks like $1 + \frac12 + \frac14 + \frac18 + \frac1{16} + ... $ for every natural $n$.
How can the sum of this (and other similar) additions be finite? I tried to understand the concept of infinite, and if a variable is finite, then it always can be bigger than a value we give for it.
In the case of the example, by each time when I increase $n$ (I make it closer to infinite), I always add a number for the sum. The bigger $n$ is, the smaller this number is, so the sum always increases in a slighty smaller amount.
I got this fact as an answer even at the university. But I still don't understand it...$n$ always increases, so the sum is also always increases! Infinite is infinite, it can be always a bigger value, no matter what we write to $n$. For an incredibly big calue, the sum must be bigger. I mean, we can prove that for a given $n$, the sum becomes bigger than 1.7, or 1.8, kor 1.95, et cetera.
For an incredibly big $n$ value, even if its bigger than that we can even display (googolplex, for example), the sum should be bigger than 2.
...at least, theoretically. I don't get it. I've heard some "real-life" examples like eating  $\frac1n$ table of chocolate every day, or the ancient Greek story (maybe from Archimedes) about the race of the turtle (or whatever), but it didn't help in understanding.
Is there anyone who can explain it on an understandable way?

An infinite number of summands is not really a problem, if the summands decrease quickly enough. For example
$$\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{k^2} = \frac{\pi^2}{6}<\infty$$
But
$$\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{k} = \infty$$
So for the convergence of series (which is what you are puzzled about), there are a lot of so-called convergence tests for the summands $a_k$ to determine, whether
$$\lim_{n\to\infty} s_n := \lim_{n\to\infty} \sum_{k=1}^n a_k$$
converges.
