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Sequences that settle down

Infinity and strange math · level 5, lesson 1 of 6

A question to start

Write down the numbers

\[1,\ \frac12,\ \frac13,\ \frac14,\ \frac15,\ \ldots\]

and keep going forever. None of them is ever 0. Every one is a little bigger than 0. And yet everybody who looks at this list says the same thing: "it is heading for 0".

Now look at this list instead:

\[-1,\ 1,\ -1,\ 1,\ -1,\ 1,\ \ldots\]

It also goes on forever, and it also keeps visiting the same numbers. Is it "heading for" anything? For 1? For \(-1\)? For 0, which is halfway between?

And one more:

\[2,\ 1.5,\ 1.333\ldots,\ 1.25,\ 1.2,\ \ldots\]

Each number is closer to 0 than the one before. Is this list heading for 0?

To answer these questions properly, we need one of the most important ideas in all of mathematics: the limit. This lesson builds it slowly, from a game.

The idea

Sequences and terms

A sequence is a list of numbers in order that goes on forever: a first number, a second, a third, and so on. Each number in the list is called a term. We usually name the terms with one letter and a small number underneath, called a subscript, that says the position:

\[a_1,\ a_2,\ a_3,\ a_4,\ \ldots\]

\(a_1\) is read "a one" and means the first term; \(a_n\) means the term in position \(n\). Often there is a rule that gives the term from its position, like \(a_n = \dfrac1n\). Then \(a_1 = 1\), \(a_2 = \dfrac12\), \(a_{10} = \dfrac{1}{10}\), and \(a_{1000} = \dfrac{1}{1000}\). You met sequences with rules in Patterns; here we only care about one question: what happens far along the list?

Picture it

Here are the first 40 terms of \(\dfrac1n\) as dots on a number line.

−0.1 0 0.1 1/3 1/2 1 window: within 0.1 of 0

The first few dots are spread out: 1, then \(\frac12\), then \(\frac13\). After that they crowd together near 0, so tightly that the drawing can hardly show them. The shaded window holds every number within 0.1 of 0 (from \(-0.1\) to \(0.1\)). The terms from the 11th on all lie inside it, because \(\dfrac1n < 0.1\) as soon as \(n > 10\).

The closeness game

Here is a game for two players that captures "heading for 0" exactly.

  1. The challenger names a closeness: "get within 0.1 of 0". Or "within 0.001". Or "within one billionth". Any positive amount, as small as they like.
  2. The defender must answer with a position in the list, such that every term from that position on is within the challenger's closeness.

For \(\dfrac1n\) the defender always wins:

Challenger says "within ..." Defender answers "from term ... on" Why
0.1 11 \(\frac1n < 0.1\) when \(n > 10\)
0.001 1001 \(\frac1n < 0.001\) when \(n > 1000\)
one millionth 1,000,001 \(\frac1n < \frac{1}{1000000}\) when \(n > 1000000\)

Whatever closeness the challenger picks, the defender takes a position past "1 divided by that closeness". Since the defender always has an answer, we say:

The sequence \(\dfrac1n\) tends to 0, or converges to 0, or has limit 0.

We write this as

\[\lim_{n \to \infty} \frac1n = 0,\]

read "the limit of 1 over \(n\), as \(n\) tends to infinity, is 0". The little arrow and \(\infty\) do not mean that \(n\) ever equals infinity. They are short for "as \(n\) grows larger and larger without end". Infinity is not a position in the list; there is no "infinitieth term" (lesson 3 explains why \(\infty\) is not a number at all).

So here is the meaning of a limit, in words:

A sequence has limit \(L\) when its terms get as close as you like to \(L\) and stay that close from some position on.

Both halves matter: "as close as you like" (the challenger can pick any closeness) and "stay there" (every term from some point on, not just some of them).

A limit can be reached, or not, from above or below

The terms do not have to reach the limit, and they do not have to come from one side.

  • \(\dfrac1n\) never reaches 0. It approaches from above (every term is bigger than 0).
  • The sequence \(\dfrac{n+1}{n}\) is \(2, \frac32, \frac43, \frac54, \ldots\) Since \(\dfrac{n+1}{n} = 1 + \dfrac1n\), each term is 1 plus a little bit, and the little bit shrinks to 0. Its limit is 1, approached from above.
0 1 2 1 5 10 15 n
  • The sequence \(5, 5, 5, 5, \ldots\) has limit 5. Every term is the limit. The defender answers "from term 1 on" every time. A constant sequence is the simplest kind that settles.
  • The sequence \(-1, \frac12, -\frac13, \frac14, -\frac15, \ldots\) jumps from one side of 0 to the other, but its distance from 0 is \(\frac1n\), which shrinks. Its limit is 0. Crossing over the limit is allowed; only the distance counts.

Sequences that do not settle

If the defender can be beaten, even by one closeness, the sequence has no limit. We say it diverges.

Example: \(-1, 1, -1, 1, \ldots\) The terms visit 1 and \(-1\) forever.

-1 0 1 1 5 10 15 n

Could the limit be 1? The challenger says "within 0.5 of 1". But every second term is \(-1\), which is 2 away from 1. However far along the defender goes, a \(-1\) comes later and breaks the rule. So 1 is not the limit. The same argument rules out \(-1\). And 0? The terms are always exactly 1 away from 0, never closer. So 0 is out too.

In fact no number works, and here is a short reason. Suppose some number \(L\) were the limit. Then from some position on, every term would be within 0.5 of \(L\). Both a 1 and a \(-1\) come after that position, so both would be within 0.5 of \(L\), which would put them within \(0.5 + 0.5 = 1\) of each other. But 1 and \(-1\) are 2 apart. That is impossible, so there is no limit.

Example: \(1, 2, 3, 4, \ldots\) The terms grow past every number. They do not settle near anything. We say this sequence tends to infinity, written \(n \to \infty\). That is a useful description, but it is still a kind of diverging: there is no number the terms close in on.

Example: \(1, 0, 1, 0, 1, 0, \ldots\) The terms land exactly on 0 infinitely often. But they keep leaving again. "Visits 0 again and again" is not the same as "settles at 0". This is why the definition says "and stays there".

A sequence cannot have two limits

Could a sequence settle at 3 and at 3.5? Picture a window of closeness 0.1 around each: one from 2.9 to 3.1, the other from 3.4 to 3.6. The windows do not overlap. If the limit were 3, the terms would eventually all be in the first window; if it were 3.5, they would eventually all be in the second. Far enough along, both would have to be true at once, and one term cannot sit in two separate windows. The same argument works for any two different numbers: pick a closeness less than half the gap between them. So a limit, if it exists, is one single number. That is what lets us write "the limit" and use an equals sign.

Worked examples

1. How far for one thousandth? For \(a_n = \dfrac1n\), from which term on is every term within 0.001 of 0?

  1. The distance from \(\frac1n\) to 0 is just \(\frac1n\).
  2. We need \(\dfrac1n < 0.001 = \dfrac{1}{1000}\), which is true exactly when \(n > 1000\).
  3. So from term 1001 on. (Term 1000 is exactly 0.001 away, which is not "less than".)

2. A rule that needs rewriting. Find the limit of \(a_n = \dfrac{2n + 3}{n + 1}\), and say from which term on every term is within 0.01 of it.

  1. Try some terms: \(a_1 = \frac52 = 2.5\), \(a_{10} = \frac{23}{11} \approx 2.09\), \(a_{100} = \frac{203}{101} \approx 2.0099\). It looks like 2.
  2. Split the top: \(2n + 3 = 2(n + 1) + 1\). So \(a_n = \dfrac{2(n+1) + 1}{n+1} = 2 + \dfrac{1}{n+1}\).
  3. Now it is clear: 2 plus a bit that shrinks to 0. The limit is 2.
  4. Distance from 2 is \(\dfrac{1}{n+1}\). It is less than 0.01 when \(n + 1 > 100\), so from term 100 on. Check: \(a_{100} = 2 + \frac{1}{101} = 2.0099\ldots\), within 0.01. ✓

3. Nines. The sequence \(0.9,\ 0.99,\ 0.999,\ 0.9999, \ldots\) has \(n\) nines in the \(n\)th term. Find its limit.

  1. The gap to 1 is \(0.1\), then \(0.01\), then \(0.001\): the \(n\)th term is \(1 - \dfrac{1}{10^n}\).
  2. The gap \(\dfrac{1}{10^n}\) shrinks to 0 (it gets ten times smaller every step), so the limit is 1.
  3. For a closeness of one millionth, \(\dfrac{1}{10^n} < \dfrac{1}{10^6}\) when \(n > 6\): from term 7 on.

Keep this example in mind. Lesson 5 shows that it is the reason \(0.999\ldots = 1\) exactly.

4. Getting closer is not enough. Does \(a_n = 1 + \dfrac1n\) tend to 0? Each term is closer to 0 than the one before: 2, 1.5, 1.33, 1.25, ...

No. Every term is bigger than 1, so every term is more than 1 away from 0. If the challenger says "within 0.5 of 0", no term at all qualifies. The terms get closer to 0, but they never get close to 0. Their limit is 1.

5. An alternating sequence that does settle. Does \(a_n = \dfrac{(-1)^n}{n}\) have a limit?

  1. \((-1)^n\) is \(-1\) when \(n\) is odd and \(1\) when \(n\) is even, so the terms are \(-1, \frac12, -\frac13, \frac14, \ldots\)
  2. The distance from 0 is always \(\frac1n\), whatever the sign.
  3. So exactly the same defender's answers as for \(\frac1n\) work. The limit is 0, even though the terms keep jumping across it.

Try it

  1. Write the first five terms of \(a_n = \dfrac{1}{2n}\). What is its limit?
  2. What is the limit of \(a_n = 3 + \dfrac1n\)? Do the terms ever reach it?
  3. For \(a_n = \dfrac1n\), from which term on is every term within 0.0001 of 0?
  4. Which of these sequences have a limit, and what is it? (a) \(7, 7, 7, \ldots\); (b) \(1, 4, 9, 16, \ldots\); (c) \(1, -\frac12, \frac13, -\frac14, \ldots\); (d) \(1, 0, 1, 0, \ldots\)
  5. Show that \(a_n = \dfrac{3n + 1}{n}\) tends to 3, and find from which term on every term is within 0.01 of 3.
  6. For \(a_n = 1 - \dfrac{1}{n^2}\), find the limit and the first position from which every term is within 0.01 of it.
  7. Find the limit of \(a_n = \dfrac{5n - 1}{n + 2}\), and from which term on every term is within 0.1 of it. (Hint: write \(5n - 1 = 5(n + 2) - 11\).)
  8. A friend says "\(0.9, 0.99, 0.999, \ldots\) tends to 1.01, because it keeps getting closer to 1.01". Use the closeness game to show the friend is wrong.
Answers
  1. \(\frac12, \frac14, \frac16, \frac18, \frac1{10}\). The limit is 0: the terms are half of \(\frac1n\), so they shrink to 0 even faster.
  2. The limit is 3. The terms are \(4, 3.5, 3.333\ldots, 3.25, \ldots\), always a little more than 3, so they never reach it. They do not need to.
  3. We need \(\frac1n < \frac{1}{10000}\), so \(n > 10000\): from term 10,001 on.
  4. (a) Limit 7 (a constant sequence). (b) No limit: the squares grow past every number (they tend to infinity). (c) Limit 0: the distance from 0 is \(\frac1n\). (d) No limit: the terms keep leaving 0 and keep leaving 1. If \(L\) were the limit, a 1 and a 0 far along would both be within 0.4 of \(L\), so within 0.8 of each other, but they are 1 apart.
  5. \(\dfrac{3n+1}{n} = 3 + \dfrac1n\), which is 3 plus a bit that shrinks to 0. The distance from 3 is \(\frac1n\), less than 0.01 when \(n > 100\): from term 101 on.
  6. The limit is 1. The distance from 1 is \(\dfrac{1}{n^2}\), which is less than \(0.01 = \frac{1}{100}\) when \(n^2 > 100\), so when \(n > 10\): from term 11 on. (Term 10 is exactly 0.01 away.)
  7. \(5n - 1 = 5(n+2) - 11\), so \(a_n = 5 - \dfrac{11}{n+2}\). The limit is 5. The distance is \(\dfrac{11}{n+2}\), less than 0.1 when \(n + 2 > 110\), so \(n > 108\): from term 109 on. Check: \(a_{109} = 5 - \frac{11}{111} = 4.9009\ldots\), which is \(0.099\ldots\) below 5. ✓
  8. Every term is less than 1, so every term is more than \(0.01\) below 1.01. The challenger says "within 0.005 of 1.01", and no term qualifies, however far along. So 1.01 is not the limit. Getting closer to 1.01 is true but not enough: the terms never get as close as you like. (The limit is 1.)

Common mistake

"The limit is the last term." An endless sequence has no last term. The limit is not a term at all (though a term may happen to equal it). It is the number the terms close in on.

"Getting closer means converging to it." The terms of \(1 + \frac1n\) get closer to 0 every single step, and also closer to \(-5\) and to \(-1000\). None of those is the limit. The limit is the one number the terms get as close as you like to, and stay. For \(1 + \frac1n\) that is 1.

Where it leads