Sequences and series

Some quantities are not computed from where they are but from what came before: each month's balance from last month's, each term from the one in front of it. A rule of that kind generates a list rather than a curve, and it needs its own machinery.

This lesson builds that machinery for the two patterns that cover most of the useful cases, and then asks the question that makes the subject interesting: what it can mean to add infinitely many numbers and get a finite answer. It assumes the exponentials of the earlier lessons and nothing else.

Sequences, explicit and recursive

A sequence is a function whose domain is the natural numbers, written $a_{1}, a_{2}, a_{3}, dots$ rather than $a (1), a (2)$ , which is notation rather than a new idea. There are two ways to specify one, and the difference matters in practice.

An explicit rule gives the $n$ -th term directly, so $a_{n} = 3 n + 2$ produces $5, 8, 11, 14$ and answers "what is the thousandth term" in one step. A recursive rule gives each term from its predecessors, so $a_{1} = 5$ with $a_{n + 1} = a_{n} + 3$ produces the same list but reaches the thousandth term only by generating the nine hundred and ninety-nine before it.

Recursive definitions are how processes actually work: a bank balance, a population census, a loan. Explicit rules are what you want for computing. Much of this lesson consists of turning a recursive description into an explicit formula, which is called finding a closed form, and it converts a thousand steps of arithmetic into one.

Arithmetic sequences and the pairing trick

A sequence is arithmetic when consecutive terms differ by a constant $d$ , the common difference. Starting from $a_{1}$ , each step adds $d$ , so after $n - 1$ steps

a_{n} = a_{1} + (n - 1) d

The count $n - 1$ rather than $n$ is where most errors live, and the check is to set $n = 1$ and confirm the formula returns $a_{1}$ .

Adding the terms gives an arithmetic series, and the trick for summing it is the one Gauss is said to have found as a schoolboy asked to add the numbers from $1$ to $100$ . Write the sum forwards and backwards, one under the other, and add columnwise. Each column gives the same total, $a_{1} + a_{n}$ , and there are $n$ columns, so twice the sum is $n (a_{1} + a_{n})$ and

S_{n} = \frac{n (a_{1} + a_{n})}{2}

For $1$ to $100$ that is $100 \times 101 / 2 = 5050$ . The formula reads as the number of terms times their average, which is a useful way to remember it and is exactly what the pairing shows.

Example. A sequence starts at $7$ and rises by $4$ each step. Find the twentieth term and the sum of the first twenty.

The twentieth term is $7 + 19 \times 4 = 83$ . The sum is $20 \times (7 + 83) / 2 = 20 \times 45 = 900$ . Check the first few by hand: $7 + 11 + 15 + 19 = 52$ , and the formula on four terms gives $4 \times (7 + 19) / 2 = 52$ .

Now you. A sequence starts at $5$ and rises by $3$ . Find the thirtieth term and the sum of the first thirty.

Answer

$a_{30} = 5 + 29 \times 3 = 92$ , and $S_{30} = 30 \times (5 + 92) / 2 = 15 \times 97 = 1455$ .

Geometric sequences and the shift trick

A sequence is geometric when consecutive terms have a constant ratio $r$ . Each step multiplies, so $a_{n} = a_{1} r^{n - 1}$ , and the exponential functions of the earlier lesson are the continuous version of the same idea.

Summing needs a different device, and it is worth seeing because the same move settles the infinite case. Write

S_{n} = a + a r + a r^{2} + dots + a r^{n - 1}

Multiply the whole thing by $r$ , which shifts every term along one place: $r S_{n} = a r + a r^{2} + dots + a r^{n}$ . Subtract the second from the first and everything in the middle cancels, leaving only the first term of one and the last of the other: $S_{n} - r S_{n} = a - a r^{n}$ . Factor and divide, which is legal provided $r \neq 1$ :

S_{n} = a \frac{1 - r^{n}}{1 - r}

Example. Sum the first ten terms of $3, 6, 12, 24, dots$

Here $a = 3$ and $r = 2$ , so $S_{10} = 3 (1 - 2^{10}) / (1 - 2) = 3 (1024 - 1) = 3069$ , and the tenth term itself is $3 \times 2^{9} = 1536$ . Note that the last term is nearly half the whole sum, which is characteristic: in a geometric series with $r = 2$ every term exceeds the total of all the terms before it.

Now you. Sum the first eight terms of $5, 15, 45, dots$

Answer

$a = 5$ , $r = 3$ , so $S_{8} = 5 (3^{8} - 1) / (3 - 1) = 5 \times 6560 / 2 = 16 {,} 400$ , and the eighth term is $5 \times 3^{7} = 10 {,} 935$ .

Adding infinitely many numbers

Now let $n$ grow without bound in that formula. The only part that depends on $n$ is $r^{n}$ , and its behaviour was settled two lessons ago: if $lvert r rvert < 1$ then repeated multiplication by $r$ drives the term toward zero. So the sum settles down to a definite value,

S_{\infty} = \frac{a}{1 - r} provided lvert r rvert < 1

and if $lvert r rvert \geq 1$ the terms do not shrink and the partial sums run away without limit. So $6 + 2.4 + 0.96 + dots$ , with $r = 0.4$ , totals $6 / 0.6 = 10$ , and no matter how many terms are taken the running total stays below $10$ and closes on it.

This retires a loose end from the first lesson. The repeating decimal $0.272727 dots$ is the series $27 / 100 + 27 / 10000 + dots$ , geometric with $a = 27 / 100$ and $r = 1 / 100$ , so its total is $(27 / 100) / (99 / 100) = 27 / 99 = 3 / 11$ , exactly the answer the shifting trick gave there. The same argument applied to $0.999 dots$ gives $(9 / 10) / (9 / 10) = 1$ . That is not an approximation and it is not a paradox: the notation $0.999 dots$ means the value the partial sums close on, and that value is $1$ .

The requirement $lvert r rvert < 1$ is essential and its failure is not subtle. Zeno's paradox of Achilles and the tortoise is a geometric series with a ratio below one, so the infinitely many gaps are crossed in a finite time and there is no paradox at all. A series with $r = 1.5$ has partial sums that grow forever, and writing down a formula's answer for it produces a number that means nothing.

What a loan costs

Compound interest and geometric series together price every fixed-payment loan, and the derivation is short enough to do rather than quote. Borrow $P$ at a monthly rate $i$ and repay a fixed amount $M$ at the end of each of $n$ months. Each payment must cancel the debt it is settling, and a payment made $k$ months from now is worth $M (1 + i)^{- k}$ today, since that sum invested now would grow to $M$ by then. Setting the total present value of the payments equal to the loan gives

P = M big [(1 + i)^{- 1} + (1 + i)^{- 2} + dots + (1 + i)^{- n} big]

The bracket is geometric with first term and ratio both $(1 + i)^{- 1}$ . Summing it and tidying gives the standard formula

M = \frac{P i}{1 - (1 + i)^{- n}}

Example. A mortgage of $200 {,} 000$ pounds at a nominal $4.2$ per cent a year runs for $25$ years with monthly payments. What is the payment, and what does the loan cost in total?

The monthly rate is $i = 0.042 / 12 = 0.0035$ and $n = 300$ . Then $(1.0035)^{- 300} = 0.350580$ , so the denominator is $0.649420$ and $M = 200000 \times 0.0035 / 0.649420 = 1077.88$ pounds a month. Over $300$ months that is $323 {,} 365$ pounds, so the interest is $123 {,} 365$ pounds, more than sixty per cent of the sum borrowed.

Now you. Find the monthly payment and the total interest on $150 {,} 000$ pounds at $5.4$ per cent over $20$ years.

Answer

$i = 0.0045$ and $n = 240$ , giving $M = 1023.38$ pounds a month. The total paid is $245 {,} 611$ pounds, so the interest is $95 {,} 611$ pounds.

The same formula run the other way prices saving. Paying $200$ a month into an account at $4$ per cent for thirty years contributes $72 {,} 000$ of capital and accumulates to $138 {,} 810$ , the difference being interest on interest.

When shrinking terms are not enough

It is tempting to conclude that a series converges whenever its terms shrink toward zero. It does not, and the standard counterexample is the harmonic series $1 + 1 / 2 + 1 / 3 + 1 / 4 + dots$ , whose terms clearly shrink to nothing.

Nicole Oresme showed around 1350 that it grows without limit, by an argument that still works. Group the terms as $1 / 3 + 1 / 4$ , then $1 / 5$ through $1 / 8$ , then $1 / 9$ through $1 / 16$ , doubling the block each time. Every term in a block is at least as large as the last one in it, and there are enough of them that each block totals at least $1 / 2$ . Infinitely many blocks each contributing at least a half cannot add to anything finite.

The growth is extraordinarily slow, which is why it fools people. The first thousand terms total $7.4855$ and the first million total $14.3927$ ; passing $10$ takes $12 {,} 367$ terms. Slow growth is still growth, and the series has no sum. Deciding which series converge, in general, is one of the main tasks of calculus, and this lesson has settled only the geometric case.

Sequences that define themselves

Some recursive sequences have no simple closed form and are still worth studying. Fibonacci's, from the Liber Abaci of 1202, is $F_{1} = F_{2} = 1$ with $F_{n + 1} = F_{n} + F_{n - 1}$ , giving $1, 1, 2, 3, 5, 8, 13, 21, dots$

Its terms grow, and the ratio of consecutive terms settles: $F_{21} / F_{20} = 17711 / 10946 = 1.618034$ , against the golden ratio $(1 + \sqrt{5}) / 2 = 1.618034$ . That is not a coincidence and it can be proved, as can a closed form for $F_{n}$ involving powers of the golden ratio, though both need tools beyond this lesson.

What the computation establishes is weaker than it looks, and this is the point on which the course now turns. Twenty terms have been checked. Nothing checked so far rules out the ratio drifting away at the two hundredth term, or the arithmetic sum formula failing at some enormous $n$ , or a pattern that holds for every case anyone has patience to test and fails after that.

Such failures are not hypothetical. The polynomial $n^{2} + n + 41$ , noted by Euler, produces a prime for every $n$ from $0$ to $39$ and then fails at $n = 40$ , where it gives $1681 = 41^{2}$ . Forty successes is a great deal of evidence and no proof at all.

So the closed forms of this lesson rest on arguments, the pairing and the shift, that were performed on a written-out list with dots in the middle. Those arguments are convincing, and being convincing is not the standard. What the standard is, and how to meet it for a claim about every natural number at once, is the last lesson.