Infinite Geometric Series

An infinite sum needs a definition

You cannot perform infinitely many additions. What you can do, for any $n$ you like, is stop early and add the first $n$ terms. That is an ordinary finite sum, and it has a name: the $n$-th partial sum, written $S_n$.

For the cake, the partial sums are

Each of these is a perfectly ordinary number, computed with finitely many additions. Nothing infinite has happened. But look at what the list is doing: the partial sums are marching toward $1$. The gap between $S_n$ and $1$ is exactly $\left(\tfrac12\right)^n$, the size of the piece you just took. The cake still on the plate always matches the last slice you cut, and that leftover shrinks as fast as you please.

That observation is what we promote into a definition.

Definition. Given an endless list of terms, form its partial sums $S_1, S_2, S_3, \ldots$ . Suppose those partial sums close in on a single number $S$, meaning they get and then stay as near to $S$ as you care to demand. Then we say the series converges, and we define its sum to be $S$. If the partial sums do not close in on any single number, the series diverges and it has no sum at all.

Three things are worth pinning down about it. First, it is a definition, not a theorem, so there is nothing to prove; we are deciding what the words will mean. Second, it agrees with ordinary addition whenever the list is finite (pad a finite list with zeros and the partial sums are eventually constant). So the definition extends arithmetic rather than contradicting it. Third, the phrase “close in on” is doing real work and has not itself been defined precisely. Making it precise is the definition of a limit, and that is the doorstep of calculus. We will lean on it in the everyday sense, and we will be scrupulous about exactly which fact we borrow.

There is a notation for the series itself. The endless sum $a + ar + ar^2 + \cdots$ is written

The $\infty$ on top is not a number, and nothing is ever substituted for it. The whole symbol is a name for the number the partial sums close in on, and it names nothing at all when they do not close in on one.

The partial sums of a geometric series

A geometric series is one whose terms have a constant ratio $r$:

The previous lesson already did the hard finite work. For $r \neq 1$,

That formula is exact for every $n$, and it is the whole reason geometric series are the ones we can handle. Now split the single fraction into two pieces, one of which contains $n$ and one of which does not:

Read that carefully, because the rest of the lesson is already visible in it. The first piece, $\tfrac{a}{1-r}$, is a fixed number: it is the same for $S_1$ and for $S_{1{,}000{,}000}$. The second piece is that same fixed number multiplied by $r^{\,n}$. So the question “what do the partial sums do?” has collapsed into a single much smaller question: what does $r^{\,n}$ do as $n$ grows?

The one fact we borrow from calculus

Everything now rests on the behavior of the powers of $r$, and here we have to be honest with you.

The Powers Fact. Let $r$ be a real number.

• If $|r| < 1$, the powers $r^{\,n}$ close in on $0$ as $n$ grows.
• If $|r| > 1$, the sizes $\left|r^{\,n}\right|$ pass every bound as $n$ grows.

Watch it happen. Multiplying repeatedly by a number smaller than $1$ in size drives the size down; multiplying repeatedly by a number larger than $1$ in size drives it up. The values below are rounded.

A table is not an argument, though, and it is easy to over-claim from one. Shrinking is not enough on its own: the numbers $1 + \tfrac1n$ shrink at every step too, and they close in on $1$, not on $0$. What is special about $r^{\,n}$ when $|r| < 1$ is that it shrinks by the same factor every step, and that is strong enough to beat any tolerance you name. Logarithms let you see it, and even compute where it happens. Suppose you want $(0.9)^n$ under $0.001$. Take logarithms of both sides, which is legal because $\log$ is increasing:

Now $\log(0.9) \approx -0.0458$ is negative, so dividing by it reverses the inequality:

From $n = 66$ onward, then, every power $(0.9)^n$ is under $0.001$, and it stays under, because the powers only shrink from there. Name a tighter tolerance and the same two lines produce a new starting point. A negative ratio needs no new work, even though you cannot take the logarithm of one: run the same two lines on $|r|$, since $\left|r^{\,n}\right| = |r|^{\,n}$. Only the size of the power is ever in question, and the sign merely flips back and forth. That is what “closes in on $0$” is going to mean.

Here is the debt, stated plainly. That computation is why the Powers Fact is believable, and it is the computation the real proof is built from, but it is not a proof. The computation is not a proof because “closes in on $0$” has not been given a precise meaning anywhere in this course. Supplying that meaning is the definition of a limit, and proving the Powers Fact from it is a calculus result you will meet later. We borrow the fact here, and we say so out loud rather than dressing a table up as a demonstration. Everything else in this lesson is airtight once you grant it.

The sum formula

The proof pays a small dividend. Writing $S = \tfrac{a}{1-r}$ for the sum, the two displayed lines say $S_n = S\left(1 - r^{\,n}\right)$, so the error you make by stopping after $n$ terms is exactly

The leftover is not merely small; it is a known number, shrinking by a factor of $|r|$ with every term you add. That is the same thing you saw in the cake, where $S = 1$ and $r = \tfrac12$. After $n$ cuts the uneaten leftover is $\left(\tfrac12\right)^n$, precisely the size of the piece you just took.

Exactly when does a geometric series converge?

The formula needed $|r| < 1$. What happens when that fails? The series has no sum, but it is worth seeing how the partial sums fail, because they fail in two genuinely different ways.

Case $r = 1$. Every term equals $a$, so $S_n = na$. (The partial-sum formula is not even available here: it divides by $1 - r = 0$.) If $a \neq 0$, the sizes $|S_n| = n|a|$ pass every bound. The partial sums run away.

Case $r = -1$. The terms are $a, -a, a, -a, \ldots$, so the partial sums are $a, 0, a, 0, \ldots$ . If $a \neq 0$ those are two different numbers, and the partial sums hop between them forever. Nothing runs away here; every partial sum is $a$ or $0$, so they stay in a bounded pen. They simply never settle on one number, and the definition demands one number. Diverging does not mean blowing up. It means not settling, and this is the case that proves the two are different.

Case $|r| > 1$. Split the partial sum again, $S_n = \tfrac{a}{1-r} - c\,r^{\,n}$ with $c = \tfrac{a}{1-r}$. The first piece is fixed. If $a \neq 0$ then $c \neq 0$, and by the Powers Fact the sizes $\left|r^{\,n}\right|$ pass every bound, so the sizes $\left|c\,r^{\,n}\right| = |c|\left|r^{\,n}\right|$ do too. A fixed number minus something unbounded is unbounded, so, when $a \neq 0$, the partial sums run away again.

The degenerate case $a = 0$. Now every term is $0 \cdot r^{\,n-1} = 0$, every partial sum is $0$, and the partial sums close in on $0$ about as convincingly as anything ever has. The all-zeros series converges to $0$ no matter what $r$ is, even $r = 10$.

That last case is exactly why you must not repeat the slogan you will hear everywhere, that “a geometric series converges exactly when $|r| < 1$.” As stated, it is false. The forward half holds: $|r| < 1$ does force convergence, for every $a$. The other half fails, and the all-zeros series with $r = 10$ is the counterexample: it converges while $|r| \geq 1$. One hypothesis repairs the statement.

Convergence criterion. Let $a \neq 0$. The infinite geometric series $a + ar + ar^2 + \cdots$ converges if and only if $|r| < 1$, and in that case its sum is $\dfrac{a}{1-r}$.

Both directions are now in hand. If $|r| < 1$, the theorem above builds the sum. If $|r| \geq 1$, one of the three cases above applies (namely $r = 1$, $r = -1$, or $|r| > 1$), and in each of them the partial sums fail to settle, so the series diverges. The hypothesis $a \neq 0$ is not decoration; it is the whole difference between a true biconditional and a false one. Many textbooks build $a \neq 0$ into the definition of a geometric sequence, so that the ratio $r = \tfrac{a_{n+1}}{a_n}$ is a sensible thing to compute at all. That convention is a perfectly good way to dodge the issue. What is not good is stating the “exactly when” with neither the convention nor the hypothesis, and then meeting the all-zeros series afterwards.

One harmless edge deserves a mention. If $r = 0$ the series reads $a + 0 + 0 + \cdots$, whose sum is plainly $a$, and the formula agrees: $\tfrac{a}{1-0} = a$. No special case is needed.

Every repeating decimal is a geometric series

Decimal notation is an infinite sum in disguise. Writing $0.4444\ldots$ means

which is geometric with $a = \tfrac{4}{10}$ and $r = \tfrac{1}{10}$. Since $|r| < 1$ the series converges, and the sum is $\tfrac{4/10}{9/10} = \tfrac49$. So a repeating decimal is a fraction, always, and now you know why: repetition is exactly what makes the digit blocks form a geometric series.

The famous case is $0.999\ldots$ . By the same reading,

Every partial sum here, $0.9$, $0.99$, $0.999$, is less than $1$. Every single one. And the sum is exactly $1$, because the sum is not one of the partial sums; it is the number they close in on. If you want to insist that $0.999\ldots$ falls short of $1$ by some positive amount $d$, name your $d$. The gap after $n$ terms is exactly $\left(\tfrac{1}{10}\right)^n$, and logarithms hand you an $n$ making that smaller than $d$. So the partial sums have already passed your claimed value, and a number cannot be smaller than the sums that are closing in on it. There is no room between $0.999\ldots$ and $1$, which is another way of saying they are two names for one number.

How far out do you have to go?

Because the error after $n$ terms is exactly $S\,r^{\,n}$, you can answer a question no table could: how many terms are enough? Logarithms turn the demand into an inequality about $n$, and you solve it.

Common mistakes