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.
A deterministic sequence - as opposed to a stochastic or random sequence.
This sequence is called the doubling sequence.
Sequence that has addition or (subtractions*) subtraction will be +(-4)
Please note that (a) this is a sequence of square numbes, and (b) the sequence starts at 22.
Arithmetic Sequence
You can use the comparison test. Since the convergent sequence divided by n is less that the convergent sequence, it must converge.
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.
(0,1,0,1,...)
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.
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.
Lexicographic preferences are not continuous because of the decreasing convergent sequence.
A sequence xn is convergent, and converges to y if given any positive number d, however small, it is possible to find a value for the index k, such thatabs(xn - y) < d for all n > k.In other words, there is a value, k, such that all elements of the sequence from xk will be closer to y than an arbitrarily small value.
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)|