How to show sequence an=a(n-1)+1/2^n is a cauchy sequence?

BY STUDYING THIS LECTURE, YOU MIGHT JUST ANSWER THIS PROBLEM ON YOUR OWN!

Any convergent sequence in any metric space is Cauchy. I think the easiest way to prove that this sequence is Cauchy is to show that it is convergent in C_b[-1,1], ie, to exhibit an element f in C_b[-1,1] and a proof that ||f_n - f|| goes to 0 as n goes to infinity (where || || denotes the supremum norm). 

It is clear that the pointwise limit of the f_n's is the function f(x) = (x^2 + 0)^(1/2) = |x|, so if the f_n's converge in the norm of C_b[-1,1] to anything it will have to be that. We just need to show that ||f_n - f|| does indeed go to 0. For this, note that for any x in [-1,1] we have 

f_n(x) - f(x) = sqrt(x^2 + 1/n) - sqrt(x^2) 
= [sqrt(x^2 + 1/n) - sqrt(x^2)] [sqrt(x^2 + 1/n) + sqrt(x^2)]/[sqrt(x^2 + 1/n) + sqrt(x^2)] 
= [(x^2 + 1/n) - x^2]/[sqrt(x^2 + 1/n) + sqrt(x^2)] 
= (1/n)/[sqrt(x^2 + 1/n) + sqrt(x^2)] 

and since x^2 >= 0 for all x we have sqrt(x^2 + 1/n) + sqrt(x^2) >= sqrt(0 + 1/n) + sqrt(0) = sqrt(1/n) and therefore 

|f_n(x) - f(x)| <= (1/n)/[sqrt(1/n)] = 1/sqrt(n) for all x in [-1,1], 

and hence ||f_n - f|| <= 1/sqrt(n) holds for all n. [If you think about this argument for a moment you will see that it actually can be used to show that ||f_n - f|| = 1/sqrt(n) for all n, but this isn't important.] Since 1/sqrt(n) goes to 0 as n goes to infinity it follows that f_n converges to f in the norm of C_b[-1,1] and hence that the sequence f_n is Cauchy in C_b[-1,1]. 

You may have noticed there is nothing special about [-1,1] here; for any closed interval [p,q] the same estimate could be used to show that f_n converges to f in the norm of C_b[p,q]. In fact the sequence f_n converges to f uniformly on all of R. [It is not easy to fit this into a metric space context because the most obvious choice of a metric space for discussing uniform convergence on R is the space C_b(R) of bounded functions on R, and while f_n - f is bounded, and hence in C_b(R), for any n, neither of the individual functions f or f_n is bounded, and hence neither f nor any of the f_n's is in C_b(R).] 

In general, when asked to prove that a sequence is Cauchy, if you can identify its limit, as we could here, the easiest thing to do is to prove that it's convergent (which implies that it's Cauchy). Proving that a sequence is Cauchy straight from the definitions is more commonly done only when the limit of the sequence cannot be explicitly identified. This isn't as rare as it sounds, e.g. when you are proving general results about convergence you generally do not have 'formulas' for the limit and must proceed in this way. For example, the usual proof of the fact that an absolutely convergent series of real numbers must be convergent does this. One uses the hypothesis that sum |a_n| is convergent to show from the definition of "Cauchy sequence" that the sequence a_1, a_1 + a_2, ..., a_1 + a_2 + ... + a_k, ... is Cauchy, and therefore convergent, and therefore that sum a_n converges. This is done because there is no explicit formula for the sum of an arbitrarily given absolutely convergent series.