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Compounding with a 100% Interest Rate


Alignments to Content Standards: F-BF.A.1.a

Task

A man knows that money in an account where interest is compounded semi-annually will earn interest faster than money in an account where interest is compounded annually. He wonders how much interest can be earned by compounding it more and more often. In this problem we investigate his question.

For ease of computation, let's suppose the man invests \$1 at a 100% interest rate. If his interest is compounded annually, his year-end balance will be: $$\begin{align} \$1 + 100 \% \cdot \$1 &= \$1 + 1.00 \cdot \$1\\ & = \$1(1+1) \\ &= \$2. \end{align}$$

If his interest is compounded semi-annually, he earns half the annual interest at mid-year, and so his mid-year balance is: $$ \begin{align} \$1 + \frac{100 \%}{2} \cdot \$1 &= \$1 + \frac{1.00}{2} \cdot \$1 \\ &= \$1 \cdot\left(1+\frac{1}{2}\right) \\ &= \$ 1.5. \end{align}$$ At year-end he earns the other half of his annual interest giving him a year-end balance of:

$$ \begin{align} \$1.5 + \frac{100 \%}{2}\$1.5 &= \$1.5 + \frac{1.00}{2}\ \cdot \$1.5 \\ &= \$ 1.5\left(1 +\frac{1}{2}\right) \\ &= \$1 \cdot\left(1+\frac{1}{2}\right)\left(1 +\frac{1}{2}\right) \\ &=\$1 \cdot\left(1+ \frac{1}{2}\right)^2\\ &= \$ 2.25. \end{align}$$

  1. Find the man's year-end balance if his interested is compounded quarterly.
  2. Write an expression which gives the man's year-end balance in terms of the number of times the interest is compounded, $n$.
  3. Now we'll investigate what happens to the year-end balance as we compound the interest more and more. This means that we want to increase the value of $n$. Complete the table below to help you see what happens to the end of year balance as $n$ becomes larger and larger. Round values to the 5th decimal place.
    $n$Year end account balance when interest is compounded $n$ times
    10
    100
    1000
    10,000
    100,000

  4. Based on the results of your table, does it appear that the man can make an unlimited amount of money off of his $1 investment if the bank compounds the interest more and more often? Explain.

IM Commentary

This task provides an approximation, and definition, of $e$, in the context of more and more frequent compounding of interest in a bank account. The approach is computational. Further progress toward finding how rapidly the approximations $\left(1 + \frac{1}{n}\right)^n$ approach $e$ could be studied by expanding and analyzing these longer and longer polynomials but going into these details is beyond the scope of the task.

This task is preliminary to F-LE Compounding Interest with a 5% Interest Rate which further develops the relationship between $e$ and compound interest.

Note that banks report account balances to the nearest cent. However, in this task we have kept several decimal places of accuracy in the account balance to clearly show changes in the account balance beyond the second decimal place.

Solution

  1. At the end of March, the man will earn one quarter of the interest giving him an account balance of

    $$\begin{align} \$1 + \frac{100 \%}{4} \cdot \$1 &= \$1 + \frac{1.00}{4} \cdot \$1 \\ &= \$1\left(1+\frac{1}{4}\right) \\ &=\$1.25. \end{align} $$

    At the end of June, the man will earn the next quarter of the interest giving an account balance of

    $$\begin{align} \$ 1.25 + \frac{100 \%}{4} \cdot \$ 1.25 &= \$ 1.25+ \frac{1.00}{4} \cdot \$ 1.25 \\ &= \$ 1.25\left(1+\frac{1}{4}\right)\\ &= \$1\left(1+\frac{1}{4}\right)^2\\ &= \$1.5625.\end{align}$$

    At the end of September, the man will earn the next quarter of the interest giving an account balance of

    $$ \begin{align} \$1.5625 + \frac{100 \%}{4}\$1.5625 &= \$1.5625 + \frac{1.00}{4}\ \cdot \$1.5625 \\ &= \$1.5625\left(1 +\frac{1}{4}\right) \\ &= \$1 \cdot\left(1+\frac{1}{4}\right)^2 \left(1 +\frac{1}{4}\right) \\ &=\$1 \cdot\left(1+ \frac{1}{4}\right)^3\\ &= \$ 1.953125. \end{align}$$

    At the end of December, the man will earn the final quarter of the interest:

    $$ \begin{align} \$1.953125 + \frac{100 \%}{4}\$1.953125 &= \$1.953125 + \frac{1.00}{4}\ \cdot \$1.953125 \\ &= \$1.953125\left(1 +\frac{1}{4}\right) \\ &= \$1 \cdot\left(1+\frac{1}{4}\right)^3 \left(1 +\frac{1}{4}\right) \\ &=\$1 \cdot\left(1+ \frac{1}{4}\right)^4\\ &= \$ 2.44140625. \end{align}$$

  2. Organizing and generalizing our previous work, we have, putting the number of times the interest is compounded in the first column and the expression showing the year end balance in the second column:

    $1$$\$1(1+\frac{1}{1})^1$
    $2$$\$1(1+\frac{1}{2})^2$
    $4$$\$1(1+\frac{1}{4})^4$
    $n$$\$1(1+\frac{1}{n})^n$

  3. Based on the work in part (b) the account balance if interest is compounded $n$ times in the year will be, in dollars, $\left(1 + \frac{1}{n}\right)^n$. The table below shows the values of this expression for the given values of $n$:

    $n$Year end account balance when interest is compounded $n$ times
    102.59374
    1002.70481
    10002.71692
    10,0002.71815
    100,0002.71827

    Unlike the values calculated in part (a), these values have been rounded to five decimal places because the full decimal would be too long to record.

  4. Based on the table, it does not appear that the man can make an unlimited amount of money off of his \$1 investment at the bank, if the bank compounds the interest more and more often. It appears that he will not make more than \$2.72. It is clear, in any case, that there are diminishing returns as the compounding frequency is increased. Going from $100$ times to $1000$ times increased the year-end balance by about one cent. Going from $1000$ to $10000$ times adds only about a tenth of a cent. And going from $10000$ times to $100000$ times yields only an extra hundredth of a cent. If this behavior continues, the account balance will not exceed $\$2.72$ no matter how often the interest is compounded.