MthT 430 Notes Chap7c Bounded Monotone Sequences Have Limits

BISHL: Bounded Increasing Sequences Have Limits

Theorem. Let {x_n}_n=1^¥ be a bounded monotone increasing sequence; i.e.

x1 £ x2 £ ¼,

and there is a number M such that for n = 1,2, ¼,

xn £ M.

Then there is a number L such that

lim n ® ¥  xn = L.

Proof using (P13-BIN): Without loss of generality, we assume that

0 £ x1 £ x2 £ ¼ £ xn £ ¼ < 1.

We will construct a binary expansion for L.

A picture is helpful!

Divide the interval [0,1) into two halves.

Is there an n₁ such that x_n1 ³ m₀ = [1/2] = 0._bin1?

If NO, let c₁ = 0, a₁ = 0 = 0._binc₁, b₁ = m₀ = a₁ + [1/2]. If YES, let c₁ = 1, a₁ = m₀ = [1/2] = 0._binc₁, b₁ = a₁ +[1/2] = 1.

In both cases, for n ³ n₁, a₁ = 0._binc₁ £ x_n £ b₁ = a₁ + [1/(2¹)] and b₁ - a₁ = [1/(2¹)].

Next Divide the interval [a₁,b₁) = [0._binc₁, a₁ + [1/(2¹)]) into two halves.

Is there an n₂ > n₁ such that x_n2 ³ m₁ = a₁ + [1/(2²)]?

If NO, let c₂ = 0, a₂ = a₁ = 0._binc₁c₂, b₂ = m₁ = a₂ + [1/(2²)]. If YES, let c₂ = 1, a₂ = m₁ = 0._binc₁c₂, b₂ = b₁ = a₂ + [1/(2²)].

In both cases, for n ³ n₂, a₂ = 0._binc₁c₂ £ x_n < b₂ = a₂ + [1/(2²)] and b₂ - a₂ = [1/(2²)].

By recursion (on k), if n_k > n_k-1, c₁, ¼, c_k, a_k = 0._binc₁¼c_k, b_k = a_k + [1/(2^k)] have been defined so that for n ³ n_k,

ak = 0.binc1¼ck £ xn < bk = ak + 1 2k ,

divide the interval [a_k,b_k) into two halves.

Is there an n_k+1 > n_k such that x_nk+1 ³ m_k = a_k + [1/(2^k+1)]?

If NO, let c_k+1 = 0, a_k+1 = a_k = 0._binc₁c₂¼c_k+1, b_k+1 = m_k = a_k+1 + [1/(2^k+1)]. If YES, let c_k+1 = 1, a_k+1 = m_k = 0._bin c₁c₂¼c_k+1, b_k+1 = b_k = a_k+1 + [1/(2^k+1)].

In both cases, n_k+1 > n_k, c₁, ¼, c_k, c_k+1, a_k+1 = 0._binc₁¼c_kc_k+1, b_k+1 = a_k+1 + [1/(2^k+1)] have been defined so that for n ³ n_k+1,

ak+1 = 0.binc1¼ck+1 £ xn < bk+1 = ak+1 + 1 2k+1 ,

Let

L = 0.binc1 ¼ck ¼ = lim k ® ¥  ak = lim k ® ¥  bk

We have that L = lim_{n ® ¥}x_n since for all k, and n > n_k,

0 £ L - xn £ bk - xn £ bk - ak £ 1 2k .

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On 24 Oct 2007, 09:41.