text: simple cauchy sequence proof? 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. 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.