# Geometric series, and why 0.999... equals 1

> What it means to add infinitely many numbers, when a geometric series has a sum, the formula a/(1 - r) with proof, and why 0.999... is exactly 1.

Series: Precalculus (https://www.monocurl.com/monocurl-essays/precalculus/)
Canonical: https://www.monocurl.com/monocurl-essays/precalculus/geometric-series/
Keywords: geometric series, sum of a geometric series, why does 0.999 equal 1, infinite series, partial sums, convergence, a over 1 minus r, zeno's paradox

[Figure 1, animation: A square of area 1 is filled by a piece of area \frac12, then \frac14, then \frac18, each piece half of what was left. The area covered after n pieces is 1 - \frac{1}{2^n}: it never equals 1, and it comes as close to 1 as you like.]

Stand one meter from a wall. Walk half of the way to it, then half of what is left, then half of what is left again, and keep going. Each step leaves some distance still to cover, so it seems you never arrive, however long you walk. And yet people reach walls all the time.

This is one of the paradoxes of Zeno of Elea, about 2500 years old, and behind it is a question with a precise answer: what should it mean to add up infinitely many numbers, here $\frac12 + \frac14 + \frac18 + \cdots$? Nobody can perform infinitely many additions, so the phrase needs a definition. Once it has one, we can prove which sums of this kind exist, find a formula for them, and settle an argument that every class has sooner or later: whether $0.999\ldots$ is equal to 1 or only very close to it.

## Half, then half of the rest

Start with the numbers. After one step you have walked $\frac12$ of a meter. After two, $\frac12 + \frac14 = \frac34$. After three, $\frac78$.

[Figure 2, steps: Walking toward a wall one meter away, each step covering half of the remaining distance. The blue part of the line is the distance walked and the red part is the distance left, which halves at every step.]

  - step: First walk half of the way: half a meter. Half a meter is left.
  - step: Then half of what is left, a quarter of a meter. You have walked 3/4 of a meter and 1/4 is left.
  - step: Then half of that, an eighth. Walked 7/8, with 1/8 left.
  - step: Three more steps, of 1/16, 1/32 and 1/64. Walked 63/64, with 1/64 left. Every step halves what is left, and no step reaches the wall.
The pattern is easier to see in what is left than in what has been walked. Each step covers half of the remaining distance, so the remaining distance is halved: $\frac12, \frac14, \frac18, \ldots$, and after $n$ steps it is $\frac{1}{2^n}$. The distance walked is the rest of the meter:

$$
\frac12 + \frac14 + \cdots + \frac{1}{2^n} = 1 - \frac{1}{2^n}.
$$

The square at the top of the page shows the same fact as areas.

**Pause and try.** How many steps does it take to get within one millimeter of the wall? Within a thousandth of a millimeter?

**Answer.** We need $\frac{1}{2^n} < 0.001$, that is $2^n > 1000$. Since $2^{10} = 1024$, ten steps are enough (and nine are not, because $2^9 = 512$). For a millionth of a meter we need $2^n > 1\,000\,000$, and $2^{20} = 1\,048\,576$, so twenty steps. Each further factor of a thousand costs only ten more steps.

So no finite number of steps reaches the wall, and yet the shortfall can be made smaller than any distance you care to name. Those two facts together are what the phrase "the sum is 1" will mean.

## What an infinite sum means

We cannot define $\frac12 + \frac14 + \frac18 + \cdots$ as the result of adding all the terms, because that process has no last step and so no result. What we do have is a list of honest finite sums: $\frac12$, $\frac34$, $\frac78$, and so on. The definition is built on that list.

**Definition.** Let $a_1, a_2, a_3, \ldots$ be an endless list of numbers. The expression $a_1 + a_2 + a_3 + \cdots$ is called a **series**. Its $n$th **partial sum** is the finite sum $s_n = a_1 + a_2 + \cdots + a_n$. The series **converges** if the sequence of partial sums $s_1, s_2, s_3, \ldots$ has a limit $S$: for every $\varepsilon > 0$ there is a whole number $N$ such that $|s_n - S| < \varepsilon$ for all $n \ge N$. In that case $S$ is called the **sum** of the series, and we write $a_1 + a_2 + a_3 + \cdots = S$. A series that does not converge **diverges**, and has no sum.

Read the clauses against the walk. The number $\varepsilon$ is a tolerance, such as one millimeter. The number $N$ is how many steps it takes to get within that tolerance, such as ten. "For all $n \ge N$" says that once inside the tolerance the partial sums stay inside. Nothing in the definition requires any partial sum to equal $S$. The sum of a series is a limit of sums, and is not itself the result of an addition.

[Figure 3, steps: The partial sums of \frac12 + \frac14 + \frac18 + \cdots plotted against the number of terms. For each tolerance \varepsilon there is a point N after which every partial sum lies within \varepsilon of 1.]

  - step: The partial sums 0.5, 0.75, 0.875, ... climb toward the dashed line at height 1 and never touch it.
  - step: Choose a tolerance of 0.1. From the fourth partial sum on, every one lies in the band between 0.9 and 1.1.
  - step: Shrink the tolerance to 0.02. Now it takes six terms, and again every later partial sum stays in the band.
Let us verify that our series converges to 1 by this definition. We have $|s_n - 1| = \frac{1}{2^n}$. One fact about powers is needed, and we will use it twice more below.

**Theorem.** **(Bernoulli's inequality.)** If $h \ge 0$ and $n$ is a whole number with $n \ge 1$, then $(1 + h)^n \ge 1 + nh$.

**Proof.** For $n = 1$ both sides are $1 + h$. Suppose the inequality holds for some $n$. Multiplying both sides by the positive number $1 + h$ gives $(1 + h)^{n+1} \ge (1 + nh)(1 + h) = 1 + (n + 1)h + nh^2 \ge 1 + (n+1)h$. So it holds for $n + 1$, and therefore for every $n \ge 1$.

With $h = 1$ this says $2^n \ge 1 + n > n$. Now let $\varepsilon > 0$ and choose a whole number $N > \frac{1}{\varepsilon}$. For every $n \ge N$ we get $|s_n - 1| = \frac{1}{2^n} < \frac1n \le \frac1N < \varepsilon$. Hence

$$
\frac12 + \frac14 + \frac18 + \cdots = 1,
$$

with the equals sign meaning exactly what the definition says.

Now back to Zeno. The walker's trouble was the belief that infinitely many steps must take forever. Suppose you walk at one meter per second. The steps then take $\frac12$ s, $\frac14$ s, $\frac18$ s, and the time used after $n$ steps is $1 - \frac{1}{2^n}$ seconds. All of the infinitely many steps happen before the clock reads 1 second, and at that moment you are at the wall. Splitting one second into infinitely many pieces does not make it longer.

## Any ratio: a shortcut and a warning

In our series each term is half of the one before. A series in which each term is a fixed multiple $r$ of the one before is called a **geometric series**, and $r$ is its **ratio**. The simplest one starts at 1:

$$
1 + r + r^2 + r^3 + \cdots
$$

There is a quick way to guess its sum. Call the sum $S$. Everything after the first term is $r$ times the whole series:

$$
S = 1 + r\,(1 + r + r^2 + \cdots) = 1 + rS,
$$

and solving $S = 1 + rS$ gives $S = \frac{1}{1 - r}$. For $r = \frac12$ this predicts $1 + \frac12 + \frac14 + \cdots = 2$, which is right, since it is 1 more than the series we just summed. For $r = \frac13$ it predicts $\frac32$, and the partial sums $1,\ 1.333,\ 1.444,\ 1.481,\ 1.494$ agree.

**Pause and try.** Apply the shortcut to $r = 2$. What does it claim about $1 + 2 + 4 + 8 + \cdots$? Is the claim believable? If it is not, which step of the argument is at fault?

**Answer.** The formula gives $S = \frac{1}{1 - 2} = -1$, so it claims $1 + 2 + 4 + 8 + \cdots = -1$. That is absurd: the partial sums $1, 3, 7, 15, \ldots$ are positive and grow without bound, so they do not approach $-1$ or anything else. Every algebraic step was valid *for a number $S$*. The fault is the first line, "call the sum $S$", which assumes the series has a sum.

So the shortcut proves only this: *if* the series converges, its sum is $\frac{1}{1-r}$. It cannot tell us when that happens. To find out we go back to the definition and look at partial sums, which are finite sums and can be handled by ordinary algebra. The same idea, multiplying by $r$ and comparing, works for them without any assumption.

## The sum of finitely many terms

Take five terms, $s_5 = 1 + r + r^2 + r^3 + r^4$. Multiplying by $r$ turns each term into the next one.

[Figure 4, steps: The partial sum s_5 = 1 + r + r^2 + r^3 + r^4 drawn as bars, for r = 0.7, and the same sum multiplied by r. Subtracting one row from the other cancels every bar except the first of the top row and the last of the bottom row.]

  - step: The five terms of the partial sum, drawn as bars. Here r = 0.7, so each bar is 0.7 times as tall as the one before it.
  - step: Multiply every bar by r. Each shrinks to the height of its right-hand neighbor, so r times the sum is the same row moved one place right, with a new last bar, r to the fifth.
  - step: Subtract the second row from the first. The bars r, r squared, r cubed and r to the fourth are in both rows, and they cancel.
  - step: Only the first bar of the top row and the last bar of the bottom row are left.
In symbols, $r s_5 = r + r^2 + r^3 + r^4 + r^5$, and subtracting gives $s_5 - r s_5 = 1 - r^5$, that is $(1 - r)\,s_5 = 1 - r^5$. Nothing here depended on there being five terms.

**Theorem.** **(Finite geometric sum.)** Let $r$ be a number with $r \ne 1$ and let $n \ge 1$ be a whole number. Then

$$
1 + r + r^2 + \cdots + r^{n-1} = \frac{1 - r^n}{1 - r}.
$$

**Proof.** Let $s_n = 1 + r + \cdots + r^{n-1}$. Then $r s_n = r + r^2 + \cdots + r^{n}$. The terms $r, r^2, \ldots, r^{n-1}$ occur in both sums, so subtracting leaves $s_n - r s_n = 1 - r^n$, that is $(1 - r)s_n = 1 - r^n$. Because $r \ne 1$ we may divide by $1 - r$.

The hypothesis $r \ne 1$ is needed only for the last division. When $r = 1$ every term is 1 and the sum of $n$ terms is $n$. Test the formula on cases we know: for $r = 2$ and $n = 3$ it gives $\frac{1 - 8}{1 - 2} = 7 = 1 + 2 + 4$, and for $r = \frac12$ it gives $s_n = 2\left(1 - \frac{1}{2^n}\right)$, which approaches 2.

## When the series converges

The partial sum is $\frac{1 - r^n}{1 - r}$, and the only part that depends on $n$ is $r^n$. For $r = \frac12$ the powers shrink toward 0 and the partial sums settle at $\frac{1}{1-r}$. For $r = 2$ the powers explode. Everything turns on the behavior of $r^n$.

**Theorem.** If $|r| < 1$, then $\lim_{n \to \infty} r^n = 0$.

**Proof.** If $r = 0$ every power is 0. Otherwise $\frac{1}{|r|} > 1$, so we can write $\frac{1}{|r|} = 1 + h$ with $h > 0$. By Bernoulli's inequality,

$$
|r^n| = \frac{1}{(1 + h)^n} \le \frac{1}{1 + nh} < \frac{1}{nh}.
$$

Given $\varepsilon > 0$, choose a whole number $N > \frac{1}{h\varepsilon}$. Then for all $n \ge N$ we have $|r^n - 0| < \frac{1}{nh} \le \frac{1}{Nh} < \varepsilon$.

**Theorem.** **(Geometric series.)** Let $a \ne 0$ and $r$ be real numbers. The series $a + ar + ar^2 + ar^3 + \cdots$ converges if and only if $|r| < 1$, and in that case

$$
a + ar + ar^2 + ar^3 + \cdots = \frac{a}{1 - r}.
$$

**Proof.** *Suppose $|r| < 1$.* By the finite sum formula the $n$th partial sum is $s_n = a\,\frac{1 - r^n}{1 - r}$, so

$$
\left|s_n - \frac{a}{1 - r}\right| = \frac{|a|}{|1 - r|}\,|r^n|.
$$

This is a fixed number $c = \frac{|a|}{|1-r|}$ times $|r^n|$. Given $\varepsilon > 0$, the previous theorem supplies an $N$ with $|r^n| < \frac{\varepsilon}{c}$ for all $n \ge N$, and then the distance above is less than $\varepsilon$. So the series converges to $\frac{a}{1-r}$.

*Suppose $|r| \ge 1$.* We first show that in any convergent series the terms have limit 0. If the partial sums $s_n$ have limit $S$ and $\varepsilon > 0$, then for large $n$ both $s_n$ and $s_{n-1}$ are within $\frac{\varepsilon}{2}$ of $S$, so the $n$th term, which is $s_n - s_{n-1}$, has size less than $\varepsilon$. In our series the terms are $ar^{k}$ with $k = 0, 1, 2, \ldots$, and $|ar^{k}| = |a|\,|r|^{k} \ge |a| > 0$ for every $k$. This covers $r = 1$, $r = -1$ and $|r| > 1$ alike. They are never smaller than $|a|$, so they do not have limit 0, and the series cannot converge.

**Key idea.** A geometric series with first term $a$ and ratio $r$ has a sum exactly when $|r| < 1$, and then the sum is $\dfrac{a}{1 - r}$: first term over one minus the ratio. The sum is by definition the limit of the partial sums $a\,\dfrac{1 - r^n}{1 - r}$, and it exists because $r^n \to 0$.

There is a picture of the formula $\frac{1}{1 - r}$. Take $0 < r < 1$ and stand squares of side $1, r, r^2, \ldots$ in a row. Their top-left corners lie on one straight line (the corner of the $k$th square is at distance $1 + r + \cdots + r^{k-1} = \frac{1 - r^k}{1 - r}$ along the base, at height $r^k$, and these points satisfy $y = 1 - (1 - r)x$). The line reaches the base at $x = \frac{1}{1-r}$, which is where the row of squares ends.

[Figure 5, animation: Squares of side 1, r, r^2, \ldots in a row, as r grows from 0.5 to 0.7. The green line through their corners meets the base where the row ends. The small orange triangle with legs 1 and 1 - r has the same shape as the large triangle with legs \frac{1}{1-r} and 1, which is the sum formula.]

**Pause and try.** The series $1 - \frac12 + \frac14 - \frac18 + \cdots$ has alternating signs. Is it geometric? Before computing, decide whether its sum should be more or less than $\frac12$, and more or less than 1. Then find it.

**Answer.** It is geometric with $a = 1$ and $r = -\frac12$, and $|r| < 1$, so it converges, to $\frac{1}{1 - (-\frac12)} = \frac{1}{3/2} = \frac23$. The partial sums are $1,\ \frac12,\ \frac34,\ \frac58,\ \frac{11}{16}, \ldots$ They jump back and forth across $\frac23$, each jump half as long as the last, so the sum lies between any two consecutive ones: between $\frac12$ and 1.

## Where it fails

Here are the partial sums for four ratios, starting from $a = 1$.

[Figure 6, image: Partial sums of 1 + r + r^2 + \cdots for four ratios. For r = \frac12 they settle at 2. For r = -1 they alternate between 1 and 0 forever. For r = 1 and r = 1.2 they grow without bound. Only the first series converges.]

The case $r = -1$ is the series $1 - 1 + 1 - 1 + \cdots$. The shortcut $S = 1 + rS$ assigns it the value $\frac12$, and in the eighteenth century several mathematicians defended that answer. By our definition the partial sums are $1, 0, 1, 0, \ldots$, which approach nothing, so the series diverges and has no sum.

The proof used the fact that the terms of a convergent series have limit 0. The converse is false: terms that shrink to 0 do not guarantee convergence. The standard counterexample is the **harmonic series** $1 + \frac12 + \frac13 + \frac14 + \cdots$. Group its terms as

$$
1 + \frac12 + \left(\frac13 + \frac14\right) + \left(\frac15 + \frac16 + \frac17 + \frac18\right) + \cdots
$$

Each bracket is more than $\frac12$: the first has two terms that are each at least $\frac14$, the next has four terms that are each at least $\frac18$, and so on. So the partial sum with $2^k$ terms is at least $1 + \frac{k}{2}$, which grows without bound. The terms of a geometric series with $|r| < 1$ do more than shrink. They shrink by a fixed factor each time, and that is why the sum can be controlled.

The next figure shows the convergence and its failure for all ratios at once.

[Figure 7, image: The partial sums s_2, s_4 and s_8 of 1 + r + r^2 + \cdots as functions of the ratio r, with the curve \frac{1}{1-r} in green. Inside the shaded strip -1 < r < 1 the partial sums close in on the green curve as terms are added. Outside it they move away.]

## Why 0.999... equals 1

What does the notation $0.999\ldots$ mean? A finite decimal such as $0.999$ is shorthand for the sum $\frac{9}{10} + \frac{9}{100} + \frac{9}{1000}$. A decimal that goes on forever cannot be a finite sum, so it is defined as a series:

**Definition.** The infinite decimal $0.d_1d_2d_3\ldots$, with digits $d_1, d_2, d_3, \ldots$, means the sum of the series $\dfrac{d_1}{10} + \dfrac{d_2}{100} + \dfrac{d_3}{1000} + \cdots$.

(Such a series always converges, because its partial sums never decrease and never exceed 1; this uses the monotone convergence theorem, that an increasing sequence which is bounded above has a limit, from the lesson on [the number $e$](@precalculus/exponential-growth-and-e).) With that definition the equation $0.999\ldots = 1$ is a statement that can be proved.

**Theorem.** $0.999\ldots = 1$.

**Proof.** By definition $0.999\ldots$ is the sum of $\frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots$, a geometric series with first term $a = \frac{9}{10}$ and ratio $r = \frac{1}{10}$. Since $|r| < 1$ it converges, and its sum is

$$
\frac{a}{1 - r} = \frac{9/10}{1 - 1/10} = \frac{9/10}{9/10} = 1.
$$

[Figure 8, image: Four number lines, each a tenfold magnification of the right-hand tenth of the one above. The partial sums 0.9, 0.99, 0.999, 0.9999 (orange) each stop short of 1 by the red gap, and the gap shrinks by a factor of 10 from one line to the next.]

The proof is three lines long, and people still resist the conclusion. The objections deserve answers.

*"0.999... is always a little less than 1."* Each partial sum $0.9$, $0.99$, $0.999$ is less than 1. But $0.999\ldots$ is none of the partial sums. It is one fixed number, the limit of the partial sums, so nothing about it is still "getting there", and a limit of numbers below 1 can equal 1.

*"The difference is $0.000\ldots1$, infinitely small but not zero."* The gaps between the partial sums and 1 are $\frac{1}{10}, \frac{1}{100}, \frac{1}{1000}, \ldots$ If $1 - 0.999\ldots$ were a positive number $d$, it would have to be smaller than every one of these gaps, because the partial sums increase toward $0.999\ldots$ and never pass it. But Bernoulli's inequality with $h = 9$ gives $10^n \ge 1 + 9n > n$, so $\frac{1}{10^n} < \frac1n$, and $\frac1n < d$ as soon as $n > \frac1d$. No positive real number is smaller than every $\frac{1}{10^n}$. And no decimal has the form "$0.000\ldots1$", because every digit of a decimal sits at some finite position, and there is no position after all of them.

*"Then one number has two different decimals."* It does. Decimals are names for numbers, and some numbers have two names, just as $\frac12$ and $\frac24$ name one fraction. Every positive number with a terminating decimal has a second one ending in nines: $0.5 = 0.4999\ldots$ and $2 = 1.999\ldots$

*"I was shown a quicker proof: $\frac13 = 0.333\ldots$, so multiplying by 3 gives $1 = 0.999\ldots$"* That argument is correct once you grant that $\frac13 = 0.333\ldots$ and that an infinite decimal can be multiplied digit by digit. Both are true, and both are facts about limits of the same kind as the one in question. It shows the equation is consistent with the rest of arithmetic; the proof is the one above.

## Worked examples

**Worked example.** **Find the sum of $6 - 3 + \frac32 - \frac34 + \cdots$.**

Each term is $-\frac12$ times the one before, so this is geometric with $a = 6$ and $r = -\frac12$. Since $|r| = \frac12 < 1$ the theorem applies:

$$
\frac{a}{1 - r} = \frac{6}{1 + \frac12} = \frac{6}{3/2} = 4.
$$

*Check.* The partial sums are $6,\ 3,\ 4.5,\ 3.75,\ 4.125$. They close in on 4 from alternate sides.

**Worked example.** **Write the repeating decimal $0.272727\ldots$ as a fraction.**

By the definition of a decimal, grouping the digits in pairs, $0.272727\ldots = \frac{27}{100} + \frac{27}{100^2} + \frac{27}{100^3} + \cdots$. This is geometric with $a = \frac{27}{100}$ and $r = \frac{1}{100}$, so

$$
0.272727\ldots = \frac{27/100}{1 - 1/100} = \frac{27/100}{99/100} = \frac{27}{99} = \frac{3}{11}.
$$

*Check.* Long division of 3 by 11 gives $0.2727\ldots$

**Worked example.** **A ball is dropped from a height of 2 meters. After each bounce it rises to $\frac35$ of its previous height. How far does it travel in total?**

The first fall is 2 m. After that, each bounce is a rise and an equal fall. The rises are $2 \cdot \frac35 = 1.2$ m, then $1.2 \cdot \frac35 = 0.72$ m, and so on: a geometric series with $a = 1.2$ and $r = \frac35$, whose sum is $\frac{1.2}{1 - 3/5} = \frac{1.2}{0.4} = 3$ m. Counting each rise twice,

$$
\text{total distance} = 2 + 2 \times 3 = 8 \text{ meters}.
$$

*Check for plausibility.* The first fall and three bounces already cover $2 + 2(1.2 + 0.72 + 0.432) = 6.704$ m, and the later bounces are small. (A real ball stops after finitely many bounces; the series is the idealized model.)

**Common mistake.** **Using $\frac{a}{1-r}$ without checking $|r| < 1$.** The formula produces a number for every $r \ne 1$: for $1 + 3 + 9 + 27 + \cdots$ it gives $\frac{1}{1 - 3} = -\frac12$. The theorem says that this series has no sum, and the number means nothing. Check the ratio first.

**Common mistake.** **Taking the wrong first term.** For $\frac13 + \frac19 + \frac1{27} + \cdots$ it is tempting to write $\frac{1}{1 - 1/3} = \frac32$. That is the sum of the series beginning with 1. Here the first term is $a = \frac13$, and the sum is $\frac{1/3}{1 - 1/3} = \frac12$. The numerator of $\frac{a}{1 - r}$ is whatever the first term is.

Stepping back: the hard step was deciding what an infinite sum is. Once the sum is defined as the limit of the partial sums, the rest is algebra on finite sums and one limit, $r^n \to 0$. The shortcut $S = 1 + rS$ is still worth knowing, as long as you remember what it assumes. The finite formula has a second life in the next lesson, where $1 - r^n = (1 - r)(1 + r + \cdots + r^{n-1})$ becomes the reason that a root of a polynomial always gives a factor.

**Check yourself.** What is the sum of $1 + \frac23 + \frac49 + \frac{8}{27} + \cdots$?
Answer: 3 It is geometric with $a = 1$ and $r = \frac23$, and $|r| < 1$, so the sum is $\frac{1}{1 - 2/3} = 3$.

**Recap.** - A series $a_1 + a_2 + \cdots$ converges to $S$ if its partial sums $s_n = a_1 + \cdots + a_n$ have limit $S$. The sum is a limit, and no partial sum need equal it.
- Finite geometric sum: $1 + r + \cdots + r^{n-1} = \frac{1 - r^n}{1 - r}$ for $r \ne 1$, proved by multiplying by $r$ and subtracting.
- If $|r| < 1$ then $r^n \to 0$ (from Bernoulli's inequality), so $a + ar + ar^2 + \cdots = \frac{a}{1 - r}$.
- If $|r| \ge 1$ and $a \ne 0$ the series diverges, because its terms do not approach 0. The shortcut $S = 1 + rS$ gives wrong answers there.
- Terms approaching 0 is necessary for convergence and not sufficient: the harmonic series diverges.
- $0.999\ldots$ is defined as the sum of $\frac{9}{10} + \frac{9}{100} + \cdots$, which is exactly 1.

## Problems

Work these on paper before opening the solutions. The first three repeat the lesson's moves, the next three combine them, and the last two go further.

**Problem 1.** Find the sum of each series, or explain why it has none. (a) $1 + \frac14 + \frac1{16} + \cdots$ (b) $5 - \frac53 + \frac59 - \cdots$ (c) $\frac23 + \frac49 + \frac8{27} + \cdots$ (d) $1 - \frac32 + \frac94 - \cdots$

**Solution.** (a) $a = 1$, $r = \frac14$: sum $\frac{1}{1 - 1/4} = \frac43$. (b) $a = 5$, $r = -\frac13$: sum $\frac{5}{1 + 1/3} = \frac{15}{4}$. (c) $a = \frac23$, $r = \frac23$: sum $\frac{2/3}{1/3} = 2$. (d) $r = -\frac32$, and $|r| = \frac32 \ge 1$, so the series diverges. Its terms $1, -1.5, 2.25, \ldots$ grow in size instead of approaching 0.

**Problem 2.** Compute $1 + 2 + 4 + \cdots + 2^{10}$ and $1 + 3 + 9 + \cdots + 3^5$ with the finite sum formula.

**Solution.** The first has 11 terms, from $2^0$ to $2^{10}$, so $n = 11$: $\frac{1 - 2^{11}}{1 - 2} = 2^{11} - 1 = 2047$. The second has 6 terms: $\frac{1 - 3^6}{1 - 3} = \frac{729 - 1}{2} = 364$. Check the second directly: $1 + 3 + 9 + 27 + 81 + 243 = 364$.

**Problem 3.** Write $0.3181818\ldots$ (the block 18 repeats) as a fraction in lowest terms.

**Solution.** Split off the part that does not repeat: $0.3181818\ldots = \frac{3}{10} + \frac{18}{1000} + \frac{18}{100000} + \cdots$. The series is geometric with $a = \frac{18}{1000}$ and $r = \frac{1}{100}$, with sum $\frac{18/1000}{99/100} = \frac{18}{990} = \frac{1}{55}$. So the number is $\frac{3}{10} + \frac{1}{55} = \frac{33}{110} + \frac{2}{110} = \frac{35}{110} = \frac{7}{22}$. Check: $7 \div 22 = 0.31818\ldots$

**Problem 4.** For which real numbers $x$ does the series $1 + 2x + 4x^2 + 8x^3 + \cdots$ converge, and what is its sum? Evaluate the sum at $x = \frac14$ and at $x = -\frac14$.

**Solution.** The series is geometric with $a = 1$ and ratio $r = 2x$. It converges exactly when $|2x| < 1$, that is $-\frac12 < x < \frac12$, and then its sum is $\frac{1}{1 - 2x}$. At $x = \frac14$ the sum is $\frac{1}{1 - 1/2} = 2$ (the series is $1 + \frac12 + \frac14 + \cdots$). At $x = -\frac14$ it is $\frac{1}{1 + 1/2} = \frac23$ (the series is $1 - \frac12 + \frac14 - \cdots$).

**Problem 5.** How many terms of $\frac12 + \frac14 + \frac18 + \cdots$ are needed for the partial sum to differ from the sum by less than $10^{-6}$? How many terms of $0.9 + 0.09 + 0.009 + \cdots$?

**Solution.** For the first series the shortfall after $n$ terms is $\frac{1}{2^n}$, and we need $2^n > 10^6$. Since $2^{19} = 524\,288$ and $2^{20} = 1\,048\,576$, we need 20 terms. For the second the shortfall is $\frac{1}{10^n}$, and $\frac{1}{10^n} < \frac{1}{10^6}$ requires $n \ge 7$: seven terms. The series with the smaller ratio converges faster, because the shortfall is multiplied by $r$ at each step.

**Problem 6.** Achilles runs at 10 meters per second and a tortoise at 1 meter per second. The tortoise starts 9 meters ahead. Zeno argues: by the time Achilles reaches the tortoise's starting point, the tortoise has moved on; by the time he reaches that point, it has moved on again; so he never catches it. Find the duration of each stage of this description, add the durations, and compare with the answer from ordinary algebra.

**Solution.** Stage 1: Achilles covers 9 m in $0.9$ s, during which the tortoise moves $0.9$ m. Stage 2: he covers $0.9$ m in $0.09$ s, and the tortoise moves $0.09$ m. Stage 3 takes $0.009$ s, and so on. Each stage is a tenth as long as the one before, so the total time is $0.9 + 0.09 + 0.009 + \cdots = \frac{0.9}{1 - 0.1} = 1$ second (this is $0.999\ldots$ seconds). By algebra: after $t$ seconds Achilles is at $10t$ and the tortoise at $9 + t$, and $10t = 9 + t$ gives $t = 1$. Zeno's description is accurate. It cuts the one second before the catch into infinitely many stages, all of which fit inside that second.

**Problem 7.** Here are three "computations" of $S = 1 - 1 + 1 - 1 + \cdots$. (i) $S = (1 - 1) + (1 - 1) + \cdots = 0$. (ii) $S = 1 + (-1 + 1) + (-1 + 1) + \cdots = 1$. (iii) $S = 1 - (1 - 1 + 1 - \cdots) = 1 - S$, so $S = \frac12$. What does the definition say, and what is the common error?

**Solution.** The partial sums are $1, 0, 1, 0, \ldots$ For any proposed limit $S$, take $\varepsilon = \frac12$: the partial sums 0 and 1 are a distance 1 apart, so they cannot both be within $\frac12$ of $S$, and both occur beyond every $N$. So the sequence has no limit and the series diverges. All three computations begin by treating $S$ as a number and then apply rules of finite arithmetic (regrouping, factoring out $-1$) to it. Those rules were never established for a symbol that stands for no number, and the three contradictory answers show what happens when they are used anyway. In (i) and (ii) the bracketing also silently selects only the even or only the odd partial sums.

**Problem 8.** Let $T_n = \frac12 + \frac{2}{4} + \frac{3}{8} + \cdots + \frac{n}{2^n}$. (a) By computing $T_n - \frac12 T_n$, show that $T_n = 2 - \frac{n + 2}{2^n}$. (b) Show that $2^n \ge \frac{n(n-1)}{2}$ for $n \ge 2$, and deduce that the series $\frac12 + \frac24 + \frac38 + \cdots$ converges to 2.

**Solution.** (a) $\frac12 T_n = \frac14 + \frac28 + \cdots + \frac{n-1}{2^n} + \frac{n}{2^{n+1}}$. Subtracting from $T_n$ term by term, the coefficient of each $\frac{1}{2^k}$ is $k - (k - 1) = 1$, so

$$
\tfrac12 T_n = \frac12 + \frac14 + \cdots + \frac{1}{2^n} - \frac{n}{2^{n+1}} = 1 - \frac{1}{2^n} - \frac{n}{2^{n+1}}.
$$

Doubling, $T_n = 2 - \frac{2}{2^n} - \frac{n}{2^n} = 2 - \frac{n+2}{2^n}$. Check: $T_2 = \frac12 + \frac12 = 1 = 2 - \frac44$. (b) $2^n$ is the number of subsets of a set of $n$ things, and $\frac{n(n-1)}{2}$ is the number of subsets with exactly two members, so $2^n \ge \frac{n(n-1)}{2}$. Hence for $n \ge 2$, $\frac{n}{2^n} \le \frac{2}{n - 1}$, and with $\frac{2}{2^n} < \frac2n$ we get $|T_n - 2| = \frac{n+2}{2^n} < \frac{2}{n-1} + \frac{2}{n}$, which is less than any given $\varepsilon$ once $n$ is large enough. So the partial sums have limit 2. The terms $\frac{n}{2^n}$ are not geometric, but the multiply-and-subtract idea reduces the sum to a geometric one.