simple cauchy sequence proof?

Question

Prove (using the definition only) that the sequence of real numbers xn:= -1 + (1/2) - (1/3) +...+ (1/n)*(-1)^n is a cauchy sequence.

A sequence xn is Cauchy if for every epsilon>0 there exists a natural number N such that for all natural numbers n,m>=N, the terms xn, xm satisfy abs(xn-xm)
Now, to prove a sequence is cauchy, you cannot assume a relationship between m and n (ie m=n+1) because the required inequality abs(xn-xm)=N.

Answer 1

Consider that every number after the nth is bounded between x_n and x_(n+1). Therefore, the maximum distance between any two terms after the nth is |x_(n+1) - x_n|, which is 1/(n+1). Since for every ε>0 there exists an n such that 1/(n+1)<ε, this sequence is cauchy.

Edit: "Now, to prove a sequence is cauchy, you cannot assume a relationship between m and n (ie m=n+1) because the required inequality abs(xn-xm)=N"

I'm not. I specifically said that _every_ term after the nth is between x_n and x_(n+1). Not that A term after the nth is, that EVERY term is. This is easily proven by induction - first we show that second term after the nth is in the interval [x_n, x_(n+1)] or [x_(n+1), x_n] (x_(n+1) trivially is, since it is an endpoint of the interval), then we suppose that x_(n+m) and x_(n+m+1) are in [x_n, x_(n+1)] or [x_(n+1), x_n], and deduce from this that since x_(n+m+2) lies between x_(n+m) and x_(n+m+1), both of which are already stipulated to lie in [x_n, x_(n+1)] or [x_(n+1), x_n], that x(n+m+2) also lies in that interval. Thus by induction, all terms lie in that interval. The maximum distance between any two numbers in an interval of length L is L, so any two elements after the nth lie closer than 1/(n+1).

I had hoped that you would be able to fill in the missing part of the proof yourself (where you show that all elements after the nth lie between x_n and x_(n+1)), on the grounds that it's intuitively obvious, but tedious to prove. What I didn't expect is for you to fail to read the part where I specifically said EVERY term after then nth and instead give me an insulting lecture about definitions I learned four years ago. I'm sorry I treated you like you were an intelligent human being.

Answer 2

Umm, Harald, you have misstated the Cauchy criterion. The criterion demands that for each t > 0 there exists an integer N such that | u_m - u_n | < t for m, n > N. in the specific occasion u_k = a million + a million/2 + ... + a million/2^ok. subsequently, if m > n u_m - u_n = a million/2^(n+a million) + ... + a million/2^m. So, enable t > 0, and choose N by way of fact the smallest constructive integer extra suitable than -log(t)/log(2). Then for m, n > N, |u_m - u_n| =a million/2^(n+a million) + ... + a million/2^m <= a million/2^(n+a million) + ... + a million/2^m + ... = a million/2^n < a million/2^N < t