3.1.8 definition

If X = (x1, x2, .., xn, …) is a sequence of real numbers and if given m natural numbers, then m-tail of X is a sequence.

Xm = (xm + n: n ϵ N) = (xm + 1, xm + 2, …)

Example:

3-tail of sequence X = (2, 4, 6, 8, 10, …, 2n, …) is sequence X3 = (8, 10, 12, …, 2n + 6 …).

3.1.9 theorem

Given X = (xn: n ϵ N) sequence of real numbers and given m ϵ N. then m-tail Xm = (xm + n: n ϵ N) of X converges if & only if X is convergent. In this case lim Xm = lim X.

Proof :

For anything p ϵ N, the p provisions in Xm are (p + m) the provisions in X. with the same thing, if q> m, then the q conditions in X are (q – m) the provisions of Xm.

Assume X is converging. Then given anything 𝛆> 0, if the X conditions for n ≥ K (𝛆) satisfy | xn – x | <𝛆, then the Xm provision for k ≥ K (𝛆) – m satisfies | xk – x | <𝛆. Thus taken Km (𝛆) = K (𝛆) – m, so Xm also converges to x.

If the Xm provision for k ≥ K (𝛆) satisfies | xk – x | <𝛆, then the X provision for n ≥ K (𝛆) + m satisfies | xn – x | <𝛆. Thus take K (𝛆) = Km (𝛆) + m.

Thus, X converges to x if & only if Xm converges to x.

3.1.10 theorem

Given (xn) sequence of real numbers and given x ϵ R. if (an) is a sequence of positive real numbers with lim (an) = 0 and if for some constant C> 0 and some m ϵ N then

| xn – x | ≤ Can for all n ≥ m,

That follows that lim (xn) = x.

Proof:

If 𝛆> 0 is given, then because lim (an) = 0, it is known that there is K = K (𝛆 / C) as n ≥ K means

an = | an – 0 | <𝛆 / C

thus it follows that if both n ≥ K and n ≥ m, then

| xn – x | ≤ Can <C (𝛆 / C) = 𝛆

Because 𝛆> 0 is arbitrary, it is concluded that x = lim (xn).