# Compound Interest Formula

I was trying to explain $$e$$ to someone and I ran across this idea. So, $$e$$ is typically defined as $$e =\lim_{n\to \infty} (1+1/n)^n$$ Which comes from the formula for compound interest $$P(t) = P_0\left(1+\frac{r}{n}\right)^{nt}.$$ I could not explain why it is that we divide by $$n$$. The typical answer is that you simply "divide" the interest rate across each compounding period, but of course means that the borrower will end up paying more. This just seems like trickery on the part of the lender. Does anybody have any idea why this formula in particular? Was it simply to trick lenders?

$$e$$ also then becomes kinda flukey. How would you justify that $$\lim_{n\to \infty} \left( 1+\frac{x}{n}\right)^n$$ is an exponential function to some base let alone to the base $$e$$?

As $$e = \lim_{n \to \infty} \left( 1 + \frac1n \right)^n,$$ and as for any real number $$r > 0$$, we have $$\lim_{n \to \infty} \frac{n}{r} = \infty = \lim_{n \to \infty} n \ \mbox{ and } \ \lim_{n \to \infty} \frac{r}{n} = 0 = \lim_{n \to \infty} \frac1n,$$ so we obtain $$\lim_{n \to \infty} \left( 1 + \frac{r}{n} \right)^n = \lim_{n \to \infty} \left( \left( 1 + \frac{r}{n} \right)^{\frac{n}{r}} \right)^r = e^r.$$

Hence we also conclude that for any real numbers $$r > 0$$ and $$t > 0$$, we have $$\lim_{n \to \infty} \left( 1 + \frac{r}{n} \right)^{nt} = \lim_{n \to \infty} \left( \left( 1 + \frac{r}{n} \right)^{\frac{n}{r}} \right)^{rt} = e^{rt}.$$

Thus if $$P_0$$ is the initial amount, $$r > 0$$ is the interest rate compounded $$n$$ times a given period of time, then after $$t$$ such periods of $$t$$, our amount will be $$P_0 \times \left( 1 + \frac{r}{n} \right)^{nt},$$ and so if our interest at rate $$r$$ is compounded continuously, then our amount after $$t$$ number of periods will be $$P_0 e^{rt}.$$

Hope this helps.

I think you mean "trick borrowers", but it doesn't trick anybody. This is simply how compound interest is normally computed, when the compounding period is shorter than one year. Does the bank "trick lenders" when it compounds interest on savings deposits this way? Imagine if your savings account paid $$4\%$$ compounded quarterly (back in the good old days), and at the end of $$3$$ months you were paid less than $$1\%$$ interest.

• I still don't get why we say 4% annual interest, when if we are compounding quaterly we end up with a 4.64% total interest at the end of the year. "Dividing the 4%" among the quarters seems very suspect, which is the usual justification for the formula $(1+\frac{0.4}{4})^4$.
– Rdrr
May 10 '20 at 19:54
• You mean $4.06\%$ There really isn't much difference. Actually, I think what happened is that banks started compounding more frequently to attract depositors. I can even remember cases where borrowers were compounding continuously. That's a come-on if you like, because the difference is very small, but it's still real. May 10 '20 at 20:01
• but $(1+\frac{0.4}{4})^4 = 1.464$. Hence you have paid 4.64% on top of the principal at the end of the year. Right?
– Rdrr
May 10 '20 at 20:04
• $4\%=.04$ not $.4$ You're using $40\%$ interest and finding that compounding it quarterly gives somewhat more that $46\%$. May 10 '20 at 20:06
• For small interest rates, indeed we have that (1+r/n)^n is pretty close to 1+r. Is that the only reason we use that formula. I don't know why we don't just give the actual rate per period.
– Rdrr
May 10 '20 at 20:12

There is a formula for case of f(x)^g(x) When f(x) tends to 1 then We can write e^((f(x)-1)*g(x)) It is applicable everywhere .. It is a odd formula but very helpful

• If anyone has any doubt then ask May 12 '20 at 5:31