((-1)^n)
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.
A convergent boundary is a deforming region where two tectonic plates or fragments move toward each other and collide. Some examples are; the forming of the Himalayas, New Zealand, and the Aleutian Islands.
The limits on an as n goes to infinity is aThen for some epsilon greater than 0, chose N such that for n>Nwe have |an-a| < epsilon.Now if m and n are > N we have |an-am|=|(am -a)-(an -a)|< or= |am -an | which is < or equal to 2 epsilor so the sequence is Cauchy.
Not always true. Eg the divergent series 1,0,2,0,3,0,4,... has both convergent and divergent sub-sequences.
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)|
You can use the comparison test. Since the convergent sequence divided by n is less that the convergent sequence, it must converge.
((-1)^n)
JUB
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.
no converse is not true
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.
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.
it is an acceptor language.it bounded with both ends
They are divided by divergent, convergent AND transform boundaries.
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.
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.