Solutions to Assignment #7. Exercises 1, 5, 9, and Statement 5 were graded for 5 points each and 5 completion points were given for the remaining problems.

Solutions to Assignment #6. Problems 3, 7, 11, and 12 were graded for 5 points each and 5 completion points were given for the remaining problems. For Problem 3, a correct proof should have involved partial sums of the series, although I did not take points off for this.

Solutions to Assignment #5. Problems 2.12.3, 2.14.7, 2.15.6, and the extra problem were graded for 5 points each and 5 completion points were given for the remaining problems. A common mistake made in exercise 2.14.7 was to not address the finitely many terms a₁,...,a_N for which the inequality (a_n)^p < a_n did not hold. Another common error occured in the extra problem. Just because we know the set of subsequential limit points S is contained in some interval [a,b] (by virtue of the fact that the sequence is bounded), it does not mean S=[a,b] (consider exercise 2.11.10). Also, a set being contained in a closed interval does not mean the set itself is closed. For example (-1,1) is contained in the closed interval [-1,1], but (-1,1) is not closed: it fails to contain -1 and 1 which are both limit points of this set. Please review the relevant definitions and talk to either Prof. Weisbart or I if you're still having trouble understanding them.

Solutions to Assignment #4. Problems 2.9.6, 2.10.7, 2.11.3, and Extra Problem 2 were graded for 5 points each and 5 completion points were given for the remaining problems. The point of part (c) of exercise 2.9.6 was to recognize that in part (a) we assumed a limit existed, while in part (b) we saw that given our starting point (x₁=1) the sequence diverges. However, if we modify our starting point we can get the sequence to converge (for example take x₁=1/2). For exercise 2.10.7, it was necessary to use induction to build this sequence. For extra problem 2, the proof I gave in class is correct but more complicated than the squeeze theorem proof. I've included both in the solutions.

Solutions to Assignment #3. Problems 2.7.5, 2.8.2(a,c,d), 2.8.6, and 2.9.3 were graded for 5 points each and 5 completion points were given for the remaining problems. The most common error are on this assignment was not fully checking the hypothesis of Theorem 9.6. To use this theorem we need to check that the limit of the denominator is non-zero.

Solutions to Assignment #2. Exercises 1.8, 1.15 and Statements 1.1 and 1.5 were graded for 5 points each and 10 completion points were given for the remaining problems. The most common error was not fully understanding what it means for a set to have the least upper bound property. A set can have this property without having an upper bound itself, we only need that subsets bounded above have a least upper bound. For example, the real numbers do not have an upper bound but they certainly have the least upper bound property (in fact one could define the reals as the smallest ordered field containing the rationals which has this property).

Solutions to Assignment #1.