Why do we say the harmonic series is divergent? If we have $\Sigma\frac{1}{n}$, why do we say it is divergent? Yes, it is constantly increasing, but after a certain point, $n$ will be so large that we will be certain of millions of digits. If we continue to let $n$ increase, we will end up with a number so large in the denominator that there will be an infinite amount of digits locked into place. What we would have left would just be an irrational number, correct? $\pi$ is an irrational number, but we still consider the value of that to be know. The common estimation of $\pi$ is 3.141592, and we can calculate it past 1,000,000 decimal places, so why can we just assume that we know the first few million places of the harmonic series, slap an irrational label on it, and call it a day? After all, the series $\Sigma\frac{1}{n^n}$ is convergent, and it basically does the same thing, it just gets there a lot faster.
I feel like argument has probably been made before, so I feel like there's probably a proof somewhere proving me wrong, if someone could point me to that.

The sum $\frac{1}{3}+\frac{1}{4}$ is $\gt \frac{1}{2}$.
The sum  $\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}$ is $\gt \frac{1}{2}$. This is easy, we have $4$ terms, each $\ge \frac{1}{8}$, and all but one $\gt \frac{1}{8}$.
The sum $\frac{1}{9}+\frac{1}{10}+\cdots+\frac{1}{16}$ is $\gt 
\frac{1}{2}$. We have $8$ terms, each $\ge \frac{1}{16}$, and all but one $\gt \frac{1}{16}$. 
The sum $\frac{1}{17}+\frac{1}{18}+\cdots +\frac{1}{32}$ is $\gt \frac{1}{2}$.
The sum of the terms $\frac{1}{33}$ to $\frac{1}{64}$ is $\gt \frac{1}{2}$. 
The sum of the terms $\frac{1}{65}$ to $\frac{1}{128}$ is $\gt \frac{1}{2}$.
And so on.
Counting in the first two terms, if we add until the term $\frac{1}{4}$ our sum is $\gt 2$. If we add up to $\frac{1}{8}$, our sum is $\gt 2.5$. Adding to $\frac{1}{16}$ puts us beyond $3$. At $32$ terms, we are beyond $3.5$. At $64$ terms, we are beyond $4$. At $256$, we are beyond $5$. At $1024$, we are beyond $6$. At $4096$, we are beyond $7$. 
Painfully slow! But if we are patient enough, after a while (a long long while) we will be beyond $10$, beyond $100$, beyond $1000$. But the universe may end first.
Remarks: On a calculator, the sum is finite! Because of roundoff, after a while we are just adding $0$. 
The answer dealt with the series $\sum \frac{1}{n}$. It turns out that for any positive $\epsilon$, the series  $\sum \frac{1}{n^{1+\epsilon}}$ converges. We can take for example $\epsilon=0.0001$. So one can say that $\sum \frac{1}{n}$ diverges extremely reluctantly, and that close neighbours converge.  
