Do I just assume that xn can be substituted for the usual xn and that xn+1 be substituted for xm and attempt to find the n,n+1 ≥ N such that | xn - xn+1 | < ε, given ε > 0?
Sorry, it's me again. Since I am trying to map {5, 10, 15, 20, ...} onto {...,-3,-2,-1,0,1,2,3} could I use the function f such that f(n) = n/5 when n is even and (-1)(n/5) when n is odd??
Our definition of convergence provided is the following:
{an} converges to A if ∀ε>0, ∃a positive integer N such that n ≥ N => | an - A | < ε.
I'm not sure how I am supposed to use this in the example above?
I'm confused. So am I simply evaluating 5n at different values for n to deduce my original set that I am trying to map to ℤ? Then that would just be {5,10,15,20,25...} so then my function would just be f(n) = 5n ?
Homework Statement : [/B]Prove that if xn is a sequence such that |xn - xn+1| ≤ (1/3n), for all n = 1,2,..., then it converges.Homework Equations : [/B]The definition of convergence.The Attempt at a Solution :[/B] I attempted to prove this by induction, so I am clearly far off the mark here...
Homework Statement : [/B]Let a = sup S. Show that there is a sequence x1, x2, ... ∈ S such that xn converges to a.Homework Equations : [/B]I know the definition of a supremum and convergence but how do I utilize these together?The Attempt at a Solution :[/B] Given a = sup S. We know that a =...
How would I go about showing that the set {5n | n = 1,2,...} is equivalent to ℤ.
I previously attempted to define a function f as f(n) = 5n/2 , when n is even AND (-1)[(5n-1)/(2)] , when n is odd. I then went on to show that the chosen function is 1-1 and onto, but my professor said this was...