Answers and Explanations

Suppose the simple interest rate is 100% for a certain period. Under simple interest alone, the money will double during that period. Under continually compounding interest, it will be multiplied by e. But what happens if interest is compounded at n equally spaced dates during that period (such as at the and of each month), rather than continually?

In the interval before the first compounding date, you are just earning simple interest on the original balance. Let's call the original balance B. The interest rate for this interval is 1/n, since the total interest rate for all n intervals is 1 (100% is just another way of writing 1). So, on the first compounding date, one gets paid interest in the amount of (1/n)B. The balance after this payment is therefore B + (1/n)B which equals (1 + (1/n)) B. In other words:

(balance at end of first interval) = (1 + (1/n)) times (balance at start of first interval)During the second interval (between the first and second compounding dates), one is again earning interest at a rate of 1/n, but this time on this new balance (1 + (1/n))B. Therefore,

= (1 + (1/n)) B.

(balance at end of second interval) = (1 + (1/n)) times (balance at start of second interval)If you do this same calculation again, you find that the balance at the end of the third interval is (1 + (1/n))^3 B and so on. The balance at the end of the nth and final interval is (1 + (1/n))^n B.

= (1 + (1/n)) times (1 + (1/n)) B

= (1 + (1/n))^2 B.

Therefore: if the interest is compounded n times during the period, the final balance is (1 + (1/n))^n times the original balance.

Compounding at more and more frequent intervals means that one is
approximating more and more closely the idea of *continual
compounding* which we talked about earlier, in which the final
balance is e times the original balance. What this means is
that

*The number e is the limit of the quantity (1 + (1/n))^n
as n goes to infinity*.

This gives a nice mathematical definition of the number e.

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