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Application: Compound Interests
The limit lim_{n→∞} (1 + 1/n)^{n} = e is closely related to compound interest. Suppose a bank pays interest on one dollar at the rate of 100% per year. If the interest is compounded n times per year the dollar will grow to (1 + 1/n) after 1/n years, to (1 + 1/n)^{k} after k/n years, and thus to (1 + 1/n)^{n} after one year. Since lim_{n→∞} (1 + 1/n)^{n} = e, one dollar will grow to e dollars if the interest is compounded continuously for one year, and to e^{1} dollars after t years. More generally, suppose the account initially has a dollars and the bank pays interest at the rate of b% per year. If the interest is compounded n times per year, the account will grow as follows: If the interest is compounded continuously the account will grow in one year to We can evaluate this limit by setting x = (100/b)n, n = (b/100)x. Thus the account grows to ae^{b/l00} dollars after one year and to ae^{bt/100} dollars after f years. Sometimes we may wish to know how rapidly a sequence grows. If two sequences approach ∞ and their quotient also approaches ∞, lim_{n→∞} a_{n} = ∞, lim_{n→∞} b_{n} = ∞, lim_{n→∞} a_{n}/b_{n} = ∞, the sequence <a_{n}> is said to grow faster than the sequence <b_{n}>. For each infinite H, both a_{H} and b_{H} are infinite. But a_{H}/b_{H} is still infinite, so a_{H} is infinite even compared to b_{H}.


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