text: # 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 $\frac{r}{n}$?
  
  • If the annual rate is $r$ and you compound $n$ times a year then each compounding should only accrue $Ar/n$ worth of interest, where $A$ is the amount that you had before that given compounding. Otherwise the nominal annual rate is different, in which case you are no longer comparing "apples to apples".
  – Ian
  Jun 3, 2016 at 5:15
  
  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). • Note that this comes out to a maximum of$5.13 a year if compounded infinitely within the year. Jun 5, 2016 at 22:48
  
  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.