Some Facts about Factorials
number n is the product of n factors, starting with n, then 1 less than n, then 2 less than n, and continuing on with each factor 1 less than the preceding one until you reach 1
Math 2260 Exam Practice Problem Solutions
X1 n=0 2n 3n+ n3: Answer: Since 3 n+ n3 >3 for all n 1, it follows that 2n 3n+ n3 < 2n 3n = 2 3 n: Therefore, X1 n=0 2n 3n+ n3 < X1 n=0 2 3 n = 1 1 2 3 = 3: Hence, the given series converges 2 Does the following series converge or diverge? Explain your answer X1 n=1 n 3n: Answer: Use the Ratio Test: lim n1 n+1 3n+1 n 3n = lim n1 n+ 1 3n+1
Math 115 Exam Solutions
1+n2 1 n = lim n→∞ n2 1+n2 = 1 Hence, the Limit Comparison Test says that the series P n 1+n2 diverges Therefore, the series P (−1) n 1+n2 converges but does not converge absolutely, so it converges condi-tionally 4 How many terms from the series X∞ n=1 1 n3 are needed to approximate the sum within 0 05? Answer: We will take the
Mathematical Induction
Example: Use mathematical induction to prove that n3 − n is divisible by 3, for every positive integer n Solution: Let P(n) be the proposition that 3 (n3 − n) –Basis: P(1) is true since 13 − 1 = 0, which is divisible by 3 –Induction: Assume P(k) holds, i e , k3 − k is divisible by 3, for an arbitrary positive integer k
Strategy for Testing Series: Solutions
MAT V1102 – 004 Solutions: page 2 of 7 8 Since ex is a strictly increasing function, e1/n ≤ e for all n ≥ 1 Hence, we have e1/n n3/2 ≤ e n3/2 Since P en−3/2 converges (it’s a p-series with p = 3/2 > 1), the comparison test
Math 116 — Practice for Exam 2
n=2 1 n √ lnn behaves as Z∞ 2 1 x √ lnx dx Z∞ 2 1 x √ lnx dx= lim b→∞ Z b 2 1 x √ lnx dx= lim b→∞ Z lnb ln2 u−1 2du= lim b→∞ 2 √ u lnb ln2 = ∞ Hence X∞ n=2 1 n √ lnn diverges 2 X∞ n=1 cos2(n) √ n3 Solution: Since 0 ≤ cos2(n) √ n3 ≤ 1 n3 2, and X∞ n=0 1 n3 2 converges by p-series test (p = 3 2 >1
Tests for Convergence of Series 1) Use the comparison test to
X1 n=1 n n2 + 1 diverges 3 X1 n=1 1 en Answer : We use the integral test with f(x) = 1=ex to determine whether this series converges or diverges To do so we determine whether the corresponding improper integral Z 1 1 1 ex dxconverges or diverges: Z 1 1 1 ex dx= lim b1 Z b 1 e xdx= lim b1 e x = lim b1 e b + e 1 = e 1: Since the integral Z
Proof by Induction
12=1, 22=4, 32=9, 42=16, (n+1)2 = n2+n+n+1 = n2+2n+1 1+3+5+7 = 42 Chapter 4 Proofs by Induction I think some intuition leaks out in every step of an induction proof — Jim Propp, talk at AMS special session, January 2000 The principle of induction and the related principle of strong induction have been introduced in the previous chapter
45 Power Series
4 5 POWER SERIES 97 4 5 Power Series A power series is a series of the form X∞ n=0 c0x n = c 0 +c1x+c2x 2 +···+c nx n +··· where x is a variable of indeterminate It can be interpreted as an
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