2
$\begingroup$

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 $\frac{r}{n}$?

$\endgroup$
  • 3
    $\begingroup$ 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". $\endgroup$ – Ian Jun 3 '16 at 5:15
6
$\begingroup$

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).

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Note that this comes out to a maximum of $5.13 a year if compounded infinitely within the year. $\endgroup$ – Simply Beautiful Art Jun 5 '16 at 22:48
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.