text: 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.\n\nEdit: "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)<epsilon must hold for ALL n,m>=N"\n\nI'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).\n\nI 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.