Math 213 Quiz # 1 Solutions

Chris Griﬃn May 1, 2008

HI EVERYBODY! It’s Chris Griﬃn here to tell you the solutions to the recent quiz # 1. HOPE YOU LIKE IT! Please see the end of the document for other general remarks about this quiz. 1. (a) This sequence converges to 3. This can be seen by factoring out an n4 on the numerator or denominator, or by L’Hospital’s Rule. Below, we use L’Hospital’s Rule: 3n4 + 1 3x4 + 1 12x3 36x2 36 lim = lim = lim = lim = lim = 3. n→∞ n4 + 2n + 1 x→∞ x4 + 2x + 1 x→∞ 4x3 + 2 x→∞ 12x2 x→∞ 12

(b) We use the squeeze theorem, and note that 0 ≤ sin2(n) ≤ 1. So sin2(n) 1 0 ≤ ≤ . n n

1 sin2(n) Now since both 0 → 0 and n → 0, we conclude that n → 0. n! 1 (c) We note that (n+2)! = (n+1)(n+2) → 0. 2. (a) This series diverges by the test for divergence: n + 1 lim = 1 6= 0. n→∞ n

1 1 P 1 (b) We use the limit comparison test with the series bn = n , and n is known to diverge since it is a p-series, with p = 1. Then

1/(n + 1) n lim = lim = 1, 0 < 1 < ∞. n→∞ 1/n n→∞ n + 1

P 1 P 1 So the limit comparison test applies, and since n diverges, we have that n+1 diverges. 1 1 Remark. A surprising number of you made a direct comparison of the form n+1 ≤ n , which P 1 is true. But since n diverges, this comparison is of no use. In a certain sense, all it would P 1 say is that n+1 ≤ ∞, which does not force it to either converge or diverge. −n 1 n 1 (c) This is a geometric series, as e = e . So since −1 < e < 1, we conclude that the given series converges. 1 P 1 (d) We use the limit comparison test with bn = n , which as mentioned before, n is a divergent p-series. We check

n/(n2 − 4n + 18) n2 lim = lim = 1. n→∞ 1/n n→∞ n2 − 4n + 18

So the limit comparison test applies, and the given series diverges. 3 n 3n P 3 n (e) We use the limit comparison test with bn = 4 = 4n . The series 4 is a convergent geometric series. Then we check that the limit comparison test applies, ﬁrst by studying the quotient an : bn n 3 n 4n−1 4 3n = n . 4n 4 − 1 Now we apply L’Hospital’s rule to see that

4n 4x (ln 4)4x lim = lim = lim = lim 1 = 1. n→∞ 4n − 1 x→∞ 4x − 1 x→∞ (ln 4)4x x→∞ Rock on!