If An and Bn are cauchy sequences prove AnBn is cauchy as we

If An and Bn are cauchy sequences, prove AnBn is cauchy as well.

Solution

If An and Bn are cauchy sequences, prove AnBn is cauchy as well.

Let epsilon > 0 , By definition there are N1, N2 belongs to N such that

| an - am | < epsilon/2 for all n,m > =N1 and | bn - bm | < epsilon/2

for all n,m > =N2

Let , N = max { N1, N2 } then for all n, m > = N2 , we have

Next, get N3, N4 belongs to N such that | an - am | < 1 for all n,m > =N2

and | bn - bm | < 1 for all n,m > =N

Set B1 = max n<= N3 { |an| +1} and B2 = max n<= N4 { |an| +1}

One readily verifies that , | an| < B1 and | bn| < B2 for all n belongs to N

This shows that all cauchy\'s sequence are bounded

Note that B1, B2 > 0 by construction . Now there are N5, N6 belongs to N

Such that | an - am | < epsilon/ (2B2) for all n,m > = N6 . Let N^t = max {N5, N6}

Then for all , n,m > = N^ t

| anbn - ambm | = | an (bn-bm) + bm( an -am) |

<= | an (bn-bm) + bm( an -am) |

= | an | . |( bn-bm) | + | bn | . |( an-am) |

< B1.epsilon/ (2B1) + B2.epsilon/ (2B2)

= epsilon

This shows that {anbn} is a Cauchy\'s Sequence