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.
0.5
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.
(0,1,0,1,...)
2,4,8,16,32,64,128....
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
0.5
(xn) is Cauchy when abs(xn-xm) tends to 0 as m,n tend to infinity.
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.
i don’t know I am Englis
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.
(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)|
Let f(x)=1/|1-x| which is continuous on the open ball yet the Cauchy sequence (xn) = (1-1/n) provides a counter example since f(xn)=n. The closed ball however does have the property thanks to its completeness.
Estrée-Cauchy's population is 321.
The population of Sauchy-Cauchy is 407.