I am proof this statement, please following:
Let (an) be a Convergent Sequence. If possible let l1 and l2 be two different limits of (an). Let excellon>0
be given. Since, (an)->l1, there exist a natural number n1 such that mod an-l1 < 1/2 excellon for all n>=n1. Since, (an)->l2, there exist a natural number n2 such that mod an-l2 < 1/2 excellon for all n>=n2.
Let m=max{n1,n2}.
Then, mod l1-l2 = mod l1-am+am-l2
<= mod am-l1 + mod am-l2
< 1/2 excellon+1/2 excellon =excellon
Therefore, mod l1-l2 < excellon, and these true for every excellon>0.
Clearly, these possible if and only if l1-l2=0. then, l1=l2.
Examples:
(i) lim n->infinity 1/n=0 (or) (1/n)->0.
Proof:
Let excellon >0 be given. Then
mod 1/n-0 < excellon
mod 1/n < excellon
1/n < excellon if n > 1/excellon
Hence, if we choose m to be any natural number such that m >1/excellon, then mod 1/n-0 < excellon for all n>m.
(ii) The Sequence ((-1)power n) is not Convergence.
Proof:
Suppose the Sequence ((-1)power n) Convergence to l.
Let excellon >0 be given there exist a natural number m such that mod(-1)power n -l < excellon for all n>=m.
Therefore, mod (-1)power m - (-1)power m+1 = mod (-1)power m -l+l -(-1)power m+1
<= mod (-1)power m -l + mod (-1)power m+1 -l
< excellon +excellon =2excellon.
But, mod (-1)power m - (-1)power m+1 =excellon
2 < 2 excellon
1 < excellon.
Hence the problem.
This is dummy text. It is not meant to be read. Accordingly, it is difficult to figure out when to end it. But then, this is dummy text. It is not meant to be read. Period.
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