∵ sequence {Xn} is bounded, so:

Xn∈{Xn}，？ M>0, when n> is N 1 (N 1∈N),

∴| xn | ≤ m hold.

∫lim(n→∞)Yn = 0

∴? ε' & gt； 0,? N2∈N, when n >; N2, there will be:

| Yn-0 | & lt； ε remains the same.

Namely: | yn | < ε'

Obviously:

| xn ||| yn | < ε' m holds, when n=max{N 1, N2}

Let ε=ε'M, then:

ε& gt； 0

| xn ||| yn | = | xnyn | < ε constant holds.

∴ There must be:

lim(n→∞) XnYn =0

Extended data:

A sequence can be regarded as a positive integer set N* or its finite set {1, 2,3, ..., n}, where {1, 2,3, ..., n} cannot be omitted.

Understanding sequence from the perspective of function is an important way of thinking. There are generally three ways to express functions, and sequence is no exception. There are usually three representations: a. list method; B. mirror image method; C. analytical methods. Among them, the analytical methods include giving the sequence with general formula and giving the sequence with recursive formula.

Functions do not necessarily have analytical formulas, and similarly, not all series have general formulas.

Derive the first n terms and formulas by inverse addition;

Sn=a 1+a2+a3？ ++ an = a 1+(a 1+d)+(a 1+2d)++[a 1+(n- 1)d]①

sn = an+an- 1+an-2 ++ a 1 = an+(an-d)+(an-2d)++[an-(n- 1)d]②

get 2sn =(a 1+an)+(a 1+an)++(a 1+an)(n)= n(a 1+an)

∴Sn=n(a 1+an)÷2。

The sum of the first n terms of arithmetic progression is equal to half of the product of the sum of the first two terms and the last two terms:

sn = n(a 1+an)÷2 = na 1+n(n- 1)d÷2

Sn=dn2÷2+n(a 1-d÷2)

There are also

a 1=2sn÷n-an

an=2sn÷n-a 1

Interestingly, S2n- 1=(2n- 1)an, S2n+1= (2n+1) an+1.