Is compound interest better than simple interest at all times????

Question

I was recently calculating compound interest for one of my questions and I came to one quite intruiging problem

so from what I understand simple interest is! when you have the same amount of interest every year so e.g Principal amount is $100$ and rate is $10\%$ then every year you get $10$.

Compound interest on other hand is when you get interest on the new value e.g. Principal amount is $100$ and rate is $10$ then first year you get $10$ interest second year you get $\$11$ interest etc and so on!

Now for the question I was working on! Principal amount is $\$11,000$ and the rate is $12\%$ compounded quarterly for 5 years.

So Simple interest would be $\text{Interest} = \text{PRT}= \$11000\cdot0.12\cdot20= \$26400$

And compound interest would be! $\text{Interest} = \$11000(1+ \frac{0.12}4)^{20}$ that is $= \$19,867.22$ HOW?? HOW is compound interest less than Simple interest while the interest each quarter for compound interest is larger than the simple interest????

Answer 1

First of all, you have two mistakes in your post.

1. The $11000\cdot 0.12\cdot 20$ calculation only finds the amount of interest accrued, while $11000\cdot (1+\frac{0.12}4)^{20}$ is the final balance of the account. The comparison between these two is meaningless. You should instead be comparing $11000(1+0.12\cdot 20)$ with $11000\cdot (1+\frac{0.12}4)^{20}$.

2. 12% is the annual interest rate (it is always annual unless otherwise stated), so the simple interest calculation should be $11000(1+0.12\cdot \color{red}{5})$.

Let's generalize. Suppose the principle is $P$, the annual interest rate is $i$, the time interval is $t$ (in years), and the compound interest divides the time interval into $n$ equal subintervals. The final account balance under simple interest is $$ P(1+it) $$ and under compound interest is $$ P\left(1+\frac{it}n\right)^n $$ The question is why the latter is always larger than the former. This follows from Bernoulli's inequality, which says $(1+x)^n\ge 1+nx$ whenever $n\ge 0$ is an integer and $x\ge 0$. Therefore, $$ \text{compound} = P\left(1+\frac{it}n\right)^n\ge P\left(1+n\cdot \frac{it}n\right)=P(1+it)=\text{simple} $$