106 - 2a Look at |(n+1)/n - (m+1)/m| = |(mn+m-mn-n)/mn| LT 1/n + 1/m; if m LT n this is less than 2/m; let H(epsilon) = 2/epsilon. (I misread the assignment, but thought I should leave this) * 6 Let S be the supremum of the sequence, by definition of supremum, *there is a point within epsilon/2 of the supremum, by definition of monotone *all subsequent members of the sequence are within epsilon/2 of the supremum, *by the triangle inequality any two subsequent members are within epsilon of *each other. 5 Choose epsilon equal .5, beyond H(epsilon) all members of the sequence differ by at most one half: since they are integers, they are equal. 9 We know Sum(r^n) = 1/(1-r) (the sum starts at n=0), hence Sum_(n=k to infinity)(r^n) = (r^k)/(1-r). We can choose k = H(epsilon) so this is less than epsilon. 109 - 2a use (1,2,3,4,5, ...) for both b use (x_n)=(n^2), (y_n)=(n) 3 Given epsilon which you want x_n to be less than, use K(1/epsilon) which assures 1/(x_n) is large; given alpha which you want 1/(x_n) to be larger than, use K(1/alpha) which assures x_n is small. 8a (n^2 + 2)^.5 GT n, hence properly diverges b n^.5 / (n^2 + 1) LT 1/n, hence converges to 0 c (n^2 + 1)^.5 / n^.5 GT n^.5, hence properly diverges d sin(n^.5) is bounded, but will get arbitrarily close to +1 and -1 infinitely often, so it diverges. 119 - 1c want |(x-1)(x+1)| LT 1/n make delta LT 1 so x+1 LT 3; make delta LT 1/3n so (1-x)(1+x) LT (1/3n)(3) = 1/n delta = min (1, 1/3n) 6 |x^2 - c^2| = |(x-c)(x+c)| LT |x-c|(a+a) [x,c LT a] therfore, let delta equal epsilon/2a so the product will be less than epsilon when |x-c| LT delta 10b |x/(1+x) - 1/2| = |(2x-1-x)/(2(1+x))| letting delta less than 1/2 the denominator is greater than 1, hence the product is less than |x-1| let delta = min (.5, epsilon) and the product will be less than epsilon. I am confused by the sequential criterion as a seperate demonstration, any sequence which converges to 1 will get within a delta neighborhood of 1 for all delta, and if delta is less than .5, the function will be within delta of 1/2 hence the sequence must converge to .5.

Sun Oct 12 1997