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Q: Examples of bounded and not convergent sequence?
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Every uniformly convergent sequence of bounded function is uniformly bounded?

The answer is yes is and only if da limit of the sequence is a bounded function.The suficiency derives directly from the definition of the uniform convergence. The necesity follows from making n tend to infinity in |fn(x)|


How do you prove that the sum of a convergent sequence divided by n will converge?

You can use the comparison test. Since the convergent sequence divided by n is less that the convergent sequence, it must converge.


What is Example of bounded sequence which is not Cauchy sequence?

((-1)^n)


A convergent sequence has a LUB or a what?

JUB


Is every cauchy sequence is convergent?

Every convergent sequence is Cauchy. Every Cauchy sequence in Rk is convergent, but this is not true in general, for example within S= {x:x€R, x>0} the Cauchy sequence (1/n) has no limit in s since 0 is not a member of S.


Any convergent sequence is a Cauchy sequence is converse true?

no converse is not true


What is a sequence which is not convergent defined as?

It could be divergent eg 1+1+1+1+... Or, it could be oscillating eg 1-1+1-1+ ... So there is no definition for a sequence that is not convergent except non-convergent.


Show that convergent sequence is bounded?

Wrong answer above. A limit is not the same thing as a limit point. A limit of a sequence is a limit point but not vice versa. Every bounded sequence does have at least one limit point. This is one of the versions of the Bolzano-Weierstrass theorem for sequences. The sequence {(-1)^n} actually has two limit points, -1 and 1, but no limit.


What are components and examples of linear bounded automata?

it is an acceptor language.it bounded with both ends


Are the major lithospheric plates bounded by diverging and converging boundaries?

They are divided by divergent, convergent AND transform boundaries.


Is the limit exists for a monotone sequence An?

If a monotone sequence An is convergent, then a limit exists for it. On the other hand, if the sequence is divergent, then a limit does not exist.


Every convergent sequence is bounded is the converse is true?

For the statement "convergence implies boundedness," the converse statement would be "boundedness implies convergence."So, we are asking if "boundedness implies convergence" is a true statement.Pf//By way of contradiction, "boundedness implies convergence" is false.Let the sequence (Xn) be defined asXn = 1 if n is even andXn = 0 if n is odd.So, (Xn) = {X1,X2,X3,X4,X5,X6...} = {0,1,0,1,0,1,...}Note that this is a divergent sequence.Also note that for all n, -1 < Xn < 2Therefore, the sequence (Xn) is bounded above by 2 and below by -1.As we can see, we have a bounded function that is divergent. Therefore, by way of contradiction, we have proven the converse false.Q.E.D.