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Math 2260 Written HW #10 Solutions 1. Consider the sequence 1 1 3 5 7 9 − , , , , , ,... . 2 5 10 17 26 37 What is the nth term in this sequence? Answer: Notice that the sequence of numerators is −1, 1, 3, 5, 7, 9, . . . which is just the odd numbers starting from −1. In fact, we can recognize that the nth term in this sequence of numerators is 2n − 3. On the other hand, the sequence of denominators is 2, 5, 10, 17, 26, 37, . . . which doesn’t look like anything in particular, but if you subtract one from each term you get 1, 4, 9, 16, 25, 36, . . . which is obviously the sequence of perfect squares. Therefore, the nth term in the sequence of denominators is n2 + 1. Putting this together, then, the nth term of the given sequence is 2n − 3 . n2 + 1 2. Consider the sequence (an )∞ n=1 where an = n! . nn Does the sequence converge or diverge? If it converges, find the limit. If it diverges, explain why. Answer: First, let me write out an slightly more explicitly: an = n · (n − 1) · (n − 2) · · · · · 3 · 2 · 1 , n · n · n · ··· · n · n · n where there are exactly n factors in both the numerator and the denominator. Re-writing slightly, I see that an = n · (n − 1) · (n − 1) · · · · · 3 · 2 1 1 1 · <1· = . n · n · n · ··· · n · 2 n n n On the other hand, an is always positive, so I have that 0 < an < 1 1 n for all n. Now, the sequence (0, 0, 0, 0, 0, 0, . . .) certainly converges to zero, and 1 =0 n→∞ n lim as well. Therefore, the sequence (an )∞ n=1 is trapped between two sequences, both converging to zero. By the Sandwich Theorem, we can conclude that lim an = 0 n→∞ as well. 2