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.

This is a definition, not a fact to prove: we are deciding what “infinite sum” will mean, and choosing it so that it agrees with ordinary addition on any finite list. The phrase “close in on” stands in for the precise idea of a limit, which calculus will define carefully. Here we use it in the everyday sense: gets close, and stays close.

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 names the series itself, the endless list of terms being added. When its partial sums close in on a single number, that same symbol also stands for that number, the sum. When they do not, the series diverges, and the symbol still names the series, just one with no sum.

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, 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, which is strong enough to beat any tolerance you name, however small. A negative ratio needs no extra care: $\left|r^{\,n}\right| = |r|^{\,n}$, so only the size of the power is ever in question, and the sign just flips back and forth.

The table is evidence, not proof. “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 are borrowing the fact here, out in the open, and calculus is where the borrowing gets paid back.

The sum formula

Grant the Powers Fact, and the rest is short. Recall the split from before:

The first piece never changes with $n$. The second piece is that same fixed number $c$ times $r^{\,n}$, and the Powers Fact says $r^{\,n}$ closes in on $0$ whenever $|r| < 1$. A fixed number times something closing in on $0$ also closes in on $0$, so the whole second piece vanishes and the partial sums close in on $\tfrac{a}{1-r} - 0 = \tfrac{a}{1-r}$. By the definition of an infinite sum, that is the series’ sum.

Convergence formula. If $|r| < 1$, then

One more thing falls out of the split for free. Writing $S = \tfrac{a}{1-r}$, the split above says $S_n = S\left(1 - r^{\,n}\right)$, so stopping after $n$ terms leaves an error of exactly

That leftover is not merely small: it is a known number, and its size, $|S - S_n| = |S|\,|r|^{\,n}$, shrinks by a factor of $|r|$ with every extra term. It is the same thing you saw in the cake, where $S = 1$ and $r = \tfrac12$: after $n$ cuts the uneaten leftover is $\left(\tfrac12\right)^n$, exactly 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 $a = 0$. The previous lesson required $a_1 \neq 0$ for a geometric sequence, so the all-zeros list $0, 0, 0, \ldots$ was never one of those. But the series $a + ar + ar^2 + \cdots$ is just an algebraic expression, defined for any real $a$ and $r$, and setting $a = 0$ is a legal thing to do to that expression even though it produces no genuine geometric sequence underneath. 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 ($r = 1$, $r = -1$, 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.

One harmless edge deserves a mention. The previous lesson also required $r \neq 0$ for a genuine geometric sequence, for the same reason as $a_1 \neq 0$: it keeps the ratio $a_{n+1}/a_n$ defined. Plug $r = 0$ into the series expression anyway and it 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 $1$, computed directly from the formula above. That can look like a contradiction: how can a sum equal $1$ when no partial sum ever does? It is not one, because the sum was never defined to be one of the partial sums. It is defined to be the number they close in on, and the gap between $S_n$ and $1$ is exactly $\left(\tfrac{1}{10}\right)^n$, a gap that shrinks past any positive amount you could name. There is no room left between $0.999\ldots$ and $1$: they are two names for one number.

Common mistakes