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.
A sequence, {xn}, is said to converge to a limit x if, given any e > 0, however small, there exists an integer k, such that abs(xk - x) < e for all n >=k. That is all value of the sequence, from xk onwrads, are no further from x than e.abs(xk - x) < e => -e < xk - x and x - xk < e which implies that xk is bounded.
It cannot be proven because it is not true.
The sequence t(n) = (-1)n is bounded: by -1 and 1 but it does not have a limit point: it oscillates.
The best answer is no answer
The Andes Mountains were formed through subduction of the oceanic plate underneath the South American plate.
American IdolAnswera tv show that starts with a in Amanda show Answera tv show that starts with a in Amanda show AnswerAmerican Dad, Alias
The title sequence of a film is the opening scene(s) that include dialogue and often action, whereas the opening credits show the cast and crew in writing, often accompanied by a song and/or some acting in the background.
Actually, it's Shanti Lowry. Wendy Raquell Robinson said so today on Wendy Williams Show. http://www.imdb.com/name/nm1283650/?ref_=nv_sr_1
They were formed by deposition and by convergent boundaries
(0,1,0,1,...)
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
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.
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.
A convergent sequence is an infinite sequence whose terms move ever closer to a finite limit. For any specified allowable margin of error (the absolute difference between each term and the finite limit) a term can be found, after which all succeeding terms in the sequence remain within that margin of error.