Why doesn't continuously compounded interest make me a zillionaire? It would seem that if I have some money and I get an interest on it every second, I'd be a zillionaire in no time. However, as the formula for the continuously compounded interest is: $A(t) = P(1 + \frac{r}{n})^{nt}$, if we go on increasing $n$, the number of times principal is compounded, there will not be much difference in the amount of money, regardless of how large the time, $t$ is.
However, not only is that extremely counter-intuitive, but what I don't get is the fact that $n$ is in the denominator inside the bracket. I understand why we raise to the power $nt$ (I think it's because if time is 3 years and we compound the money semi-annually, then deposit wil be made 6 times). But why is it not the case inside the brackets? Shouldn't the term be $rn$ and not $\frac{r}{n}$?
 A: Let's say you have \$100 invested at a 5% annual interest rate, compounded yearly. After a year you get \$5 in returns. If instead it's compounded every six months, you get \$2.50 after 6 months, not \$5. That's the definition of an "annual" interest rate, independent of how often it's compounded. You still get a little more if it's compounded more quickly, but not dramatically more. E.g. if your \$100 were compounded twice a year at a 5% annual interest rate like I said before, you'd get back \$2.50 after the first six months, but then \$2.56 after the second (since now you're accruing interest on \$102.50 instead of just \$100).
A: The reason why that term in the brackets is $\frac rn$ has been well explained, but I think the part about why it's counterintuitive that your money won't go up without bound in the case of continuous compounding is worth a closer look.
Start with $A(t) = P(1 + \frac{r}{n})^{nt}$ and put $n=kr$. The equation becomes $A(t) = P(1 + \frac{1}{k})^{k(rt)}$.
Now, when the intervals between compounding become "infinitesimal", you're considering the limit of that expression as $k \to \infty$:
$A(t) = \lim_{k \to \infty} P(1 + \frac{1}{k})^{k(rt)} = P(\lim_{k \to \infty} {(1 + \frac{1}{k})^k})^{rt} = e^{rt}$
where $e$ is the base of the natural logarithm.
You should now be able to see that even with "infinite" compoundings, you will still have a finite gain.
