## Why study sequences and series?

A sequence is simply a list of numbers, and a series is the sum of a list of numbers. So any time you have data arranged in a list, you may require methods from sequences and series to analyze the data.

For example, suppose you take out a small student loan for $10,000. When the loan is in repayment, it will accrue 6% annual interest, compounded each month. Suppose that the monthly payments were fixed at$300 by the lender. How long will it take to pay off the loan? How much interest will you end up paying?

To answer these questions quickly, we would have to know about certain loan formulas. However we can still answer the questions by finding out exactly what is still left to pay after each month. That is, we will examine the sequence of balances.

Let’s work out the first month carefully. First of all, the bank will tack on interest. Each month, the bank charges 0.5% interest $(6\% \div 12)$, or as a decimal: 0.005.  et $I_n$ stand for the interest payment in month $n$.

$I_1 = \10,\!000 \times 0.005 = \50$

Therefore out of the first $300, only$250 will go towards paying off the loan. Let $A_n$ be the balance after month $n$. By convention, we write $A_0 = 10,\!000$ for the initial amount of the loan (let’s not worry about the dollar signs going forward). So after the first month, the new balance would be:

$A_1 = 10,\!000 - 250 = 9,\!750$

To compute the balance after the second month, we repeat the same steps. The new interest payment is:

$I_2 = 9750 \times 0.005 = 48.75$

That would leave $300-48.75=251.25$ for paying off the loan. Altogether then, the new balance will be:

$A_2 = 9750 - 251.25 = 9498.75$

But now in order to automate the process, let’s write everything in terms of $I_n$ and $A_n$. First, the interest payment is always equal to the previous balance times 0.005.

$I_n = A_{n-1} \times 0.005$

Then the new payment is $300 - I_n$. This leads to the formula for the new balance (which we can simplify algebraically):

\begin{align}A_n& = A_{n-1} - (300 - I_n) \\[1mm] &= A_{n-1} - 300 + I_n \\[1mm] &= A_{n-1} - 300 + A_{n-1} \times (0.005)\\[1mm] &= (1.005)A_{n-1} - 300 \end{align}

Although it seems complicated at first, the formula for $A_n$, which is called a recursive definition for the sequence, helps us to compute all of the balances very quickly. The reason is that each balance calculation is found by a simple formula from the previous. In fact, spreadsheets are very good at this kind of work. Here is a table of the first six months of balances and interest payments.

 Month Balance Interest Paid 0 $10,000 – 1$9750 $50 2$9498.75 $48.75 3$9246.24 $47.49 4$8992.47 $46.23 5$8737.43 $44.96 6$8481.12 \$43.69

Try filling out more of the table. You may find that after about 3 years, the student loan will be paid off.

How much interest did you pay over the life of the loan? In this module you will also learn about sums of sequences, or series. Just add the terms of the series $I_1 + I_2 + I_3 + \cdots$ in order to figure out how much total interest was paid. That 6% interest really adds up, doesn’t it?