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.
((-1)^n)
0.5
(0,1,0,1,...)
A= 1.5113 b= 4.2
no converse is not true
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.
The Cauchy constant, also known as the Cauchy sequence property, tells us that a sequence is convergent if it is a Cauchy sequence. This means that for any arbitrarily small positive number ε, there exists an index after which all elements of the sequence are within ε distance of each other. It is a key property in the study of convergence in mathematics.
((-1)^n)
A compact metric space is not necessarily complete. Compactness only guarantees that every sequence in the space has a convergent subsequence, while completeness requires that every Cauchy sequence converges to a point in the space.
Consider the sequence (a_i) where a_i is pi rounded to the i_th decimal place. This sequence clearly contains only rational numbers since every number in it has a finite decimal expansion. Furthermore this sequence is Cauchy since a_i and a_j can differ at most by 10^(-min(i,j)) or something which can be made arbitrarily small by choosing a lower bound for i and j. Now note that this sequence converges to pi in the reals, so it can not converge in the set of rational numbers. Therefore the rational numbers allow a non-convergent Cauchy sequence and are thus by definition not complete.
0.5
(xn) is Cauchy when abs(xn-xm) tends to 0 as m,n tend to infinity.
You can use the comparison test. Since the convergent sequence divided by n is less that the convergent sequence, it must converge.
i don’t know I am Englis
JUB
(0,1,0,1,...)