331- 1 a) 1/(((n+1)(n+2)) .lt. 1/n^2, so converges (use integral test if necessary) b) n/((n+1)(n+2)) .gt. .5(n+2)/((n+2)(n+2)) for n .gt. 2, and sum (1/n) diverges. c) 2^(-1/n) approaches 2^0 = 1, so the summation diverges. d) n/2^n The summation will converge by either the root or ratio test. lim (n goto infinity) ((n+1)/(2^(n+1)))/(n/2^(n)) = .5 4 a) e .gt. 2, so the ratio is .lt. 1 and the series converges. b) n is eventually .gtr. e, so the series diverges. c) by definition this is 1/n, which we know provides a divergent series. d) log(n)e^-(n^.5) .lt. ne^-(n^.5) .lt. 1/n^2 for n suff large (use L'Hopital on n^3 / e^-(n^.5) ), so convergent. e) ratio test provides n/e, so divergent f) ratio test provides (n+1)/e^(2n+1), so convergent. 10 |x_n+1| .lt. r|x_n|, and by induction |x_n+i| .lt. r^i |x_n| summing |x_n|r^i from 1 to infinity provides the result. 336- 1 All are alternating, a, b, and d converge since terms decrease to 0, c diverges since terms do not go to 0. a is absolutely convergent since n^2+1 .gt. n^2, b and d are not absolutely convergent since both are essentially 1/n or worse. 7 Since it is alternating, must shw (log n)^p / n^q goes to zero. consider log(n) / n^(q/p) use L'Hopital to show this goes to 0, hence so does its p-th power. 9 Apply Dirichlet's test x_n = e^(-nt), y_n = a_n.

Sun Jan 25 1998