What Is Mathematics? An Elementary Approach to Ideas and Methods, 2nd Edition (1996)

§1. EXAMPLES OF LIMITS

1. General Remarks

In many cases the convergence of a sequence a_n can be proved by an argument of the following sort. We find two other sequences, b_n and c_n, whose terms have a simpler structure than those of the original sequence, and such that

for every n. Then if we can show that the sequences b_n and c_n both converge to the same limit α, it follows that a_n also converges to the limit α. We shall leave the formal proof of the statement to the reader.

It is clear that applications of this procedure will involve the use of inequalities. It is therefore appropriate to recall a few elementary rules which govern arithmetical operations with inequalities.

1. If a > b, then a + c > b + c (any number may be added to both sides of an inequality).

2. If a > b and the number c is positive, then ac > bc (an inequality may be multiplied by any positive number).

3. If a < b, then – b < – a (the sense of an inequality is reversed if both sides are multiplied by –1). Thus 2 < 3 but –3 < –2.

4. If a and b have the same sign, and if a < b, then 1/a > 1/b.

5. | a + b | ≤ | a | + | b |.

2. The Limit of qⁿ

If q is a number greater than 1, the sequence qⁿ will increase beyond any bound, as does the sequence 2, 2², 2³, · · · for q = 2. The sequence “tends to infinity” (see p. 294). The proof in the general case is based on the important inequality (proved on p. 15)

where h is any positive number. We set q = 1 + h, where h > 0; then

If k is any positive number, no matter how large, then for all n > k/h it follows that

qⁿ > nh> k;

hence qⁿ → ∞.

If q = 1, then the members of the sequence qⁿ are all equal to 1, and 1 is therefore the limit of the sequence. If q is negative, then qⁿ will alternate between positive and negative values, and will have no limit if q ≤ –1.

Exercise: Give a rigorous proof of the last statement.

On page 64 we showed that if –1 < q < 1, then qⁿ → 0. We may give another and very simple proof of this fact. First we consider the case where 0 < q < 1. Then the numbers q, q², q³ · · · form a monotone decreasing sequence bounded below by 0. Hence, according to page 295, the sequence must approach a limit: qⁿ → a. Multiplying both sides of this relation by q we obtain qⁿ⁺¹ → aq.

Now qⁿ⁺¹ must have the same limit as qⁿ, since the name, n or n + 1, of the increasing exponent, does not matter. Hence aq = a, or a(q – 1) = 0. Since 1 – q ≠ 0, this implies that a = 0.

If q = 0, the statement qⁿ → 0 is trivial. If –1 < q < 0, then 0 < | q | < 1; hence | qⁿ | = | q |ⁿ → 0 by the preceding argument From this it follows that always qⁿ → 0 for | q | < 1. This complete the proof.

Exercises: Prove that for n → ∞:

1) (x²/1 + x²)ⁿ → 0;

2) (x/1 + x²)ⁿ → 0;

3) (x³/4 + x²)ⁿ tends to infinity for x > 2, to 0 for | x | < 2.

3. The Limit of

The sequence , i.e. the sequence , has the limit 1 for any fixed positive number p:

(By the symbol we mean, as always, the positive nth root. For negative numbers p there are no real nth roots when n is even.)

To prove the relation (3), we first suppose that p > 1; then will also be greater than 1. Thus we may set

where h_n is a positive quantity depending on n. The inequality (2) then shows that

p = (1 + h_n)ⁿ > nh_n.

On dividing by n we see that

0 < h_n < p/n.

Since the sequences b_n = 0 and c_n = p/n both have the limit 0, it follows by the argument of Article 1 that h_n also has the limit 0 as n increases, and our assertion is proved for p > 1. Here we have a typical instance where a limiting relation, in this case h_n → 0, is recognized by enclosing h_nbetween two bounds whose limits are more easily obtained.

Incidentally, we have derived an estimate for the difference h_n between and 1; this difference must always be less than p/n.

If 0 < p < 1, then , and we may set

where h_n is again a positive number depending on n. It follows that

so that

From this we conclude that h_n tends to 0 as n increases. Hence, since , it follows that .

The equalizing effect of nth root extraction, which tends to push every positive number towards 1 as n increases, is even strong enough to do this in some cases if the radicand does not remain constant. We shall prove that the sequence tends to 1, i.e. that

as n increases. By a little device this can again be shown to follow from the inequality (2). Instead of the nth root of n, we take the nth root of . If we set , where k_n is a positive number depending on n, then the inequality yields , so that

Hence

The right side of this inequality tends to 1 as n increases, so that must also tend to 1.

4. Discontinuous Functions as Limits of Continuous Functions

We may consider limits of sequences a_n when a_n is not a fixed number but depends on a variable x: a_n = f_n(x). If this sequence converges as n → ∞, then the limit is again a function of x,

f(x) = lim f_n(x).

Such representations of functions f(x) as limits of others are often useful in reducing “higher” functions f(x) to elementary functions f_n(x).

This is true in particular of the representation of discontinuous functions by explicit formulas. For example, let us consider the sequence for every n, so that f_n(x) → 1/2. For | x | > 1 we have x²ⁿ → 0, and hence f_n(x) → 1, while for | x | > 1 we have x²ⁿ → ∞, and hence f_n(x) → 0. Summarizing:

Here the discontinuous function f(x) is represented as the limit of a sequence of continuous rational functions.

Another interesting example of a similar character is given by the sequence

For x = 0 all the values f_n(x) are zero, and therefore f(0) = lim f_n(0) = 0. For x ≠ 0 the expression 1/(1 + x²) = q is positive and less than 1; our results on geometrical series guarantee the convergence of f_n(x) for n → ∞. The limit, i.e. the sum of the infinite geometrical series, is , which is equal to 1 + x². Thus we see that f_n(x) tends to the function f(x) = 1 + x² for x ≠ 0, and to f(x) = 0 for x = 0. This function has a removable discontinuity at x = 0.

*5. Limits by Iteration

Often the terms of a sequence are such that a_n+1 is obtained from a_n by the same procedure as a_n from a_n–1; the same process, indefinitely repeated, produces the whole sequence from a given initial term. In such cases we speak of a process of “iteration.”

For example, the sequence

has such a law of formation; each term after the first is formed by taking the square root of 1 plus its predecessor. Thus the formula

defines the whole sequence. Let us find its limit. Obviously a_n is greater than 1 for n > 1. Furthermore a_n is a monotone increasing sequence, for

Hence whenever a_n > a_n–1 it will follow that a_n+1 > a_n. But we know that , from which we conclude by mathematical induction that a_{n + 1} > a_n for all n, i.e. that the sequence is monotone increasing. Moreover it is bounded; for by the previous results we have

By the principle of monotone sequences we conclude that for n → ∞ a_n → a, where a is some number between 1 and 2. We easily see that a is the positive root of the quadratic equation x² = 1 + x. For as n → ∞ the equation becomes a² = 1 + a. Solving this equation, we find that the positive root is . Thus we may solve this quadratic equation by an iteration process which gives the value of the root with any degree of approximation if we continue long enough.

We can solve many other algebraic equations by iteration in a similar way. For example, we may write the cubic equation x³ − 3x + 1 = 0 in the form

We now choose any value for a₁, say a₁ = 0, and define

obtaining the sequence a₂ = 1/3 =.3333..., a₃, = 9/26 =.3461..., a₄ = 676/1947 =.3472..., etc. It may be shown that the sequence a_n obtained in this way converges to a limit a =.3473 … which is a solution of the given cubic equation. Iteration processes such as this are highly important both in pure mathematics, where they yield “existence proofs,” and in applied mathematics, where they provide approximation methods for the solution of many types of problems.

Exercises on limits. For n → ∞: