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Berkeley MATH 1B - Math 1B Section 107 Quiz

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Math 1B Section 107 Quiz #7Thursday, 11 October 2007GSI: Theo Johnson-Freydhttp://math.berkeley.edu/∼theojfName:1. True or False (1 pt each) For each of the following statements, decide if it is trueor false. You do not need to show work: I will grade only your answers.(a) Let’s say 0 ≤ an≤ bn, andP∞n=1an= A andP∞n=1bn= B. If A = B, thenan= bnfor every n.TRUE. The first inequality guarantees thatP(bn− an) = B − A = 0 is a sumof nonnegative terms, but the only way to add up nonnegative numbers and get0 is if each of the numbers is itself 0.(b) If 0 ≤ an≤ f(n), where f(x) is a continuous decreasing function on x ∈ [1, ∞),such thatR∞1f(x) dx converges, thenP∞n=1anconverges.TRUE. This is a combination of the comparison and integral tests.(c) If 0 ≤ an≤ π/2 andP∞n=1sin(an) diverges, thenP∞n=1andiverges.TRUE. This follows from the comparison test, since sin(x) < x if x > 0, and soan> sin(an).1For the next two questions, use either the Limit Comparison Test or the IntegralTest to determine if the series converges or diverges. Be sure to check that the seriessatisfies the conditions necessary for the test.2. (3 pts)∞Xn=1arctan(n)n2− ln(n)We recall that limn→∞arctan(n) = π/2. Then comparing with 1/n2(which con-verges) giveslimn→∞arctan(n)/(n2− ln n)1/n2= limn→∞arctan(n)11 − (ln(n)/n2)=π2· 1which is more than 0 and less than ∞. So sinceP1/n2converges, so must our seriesParctan(n)/(n2− ln n).3. (4 pts)∞Xn=31n ln n ln(ln n)Recognizing this as something integrable, we check the conditions for the integral test:• 1/(x ln x ln(ln x)) is always positive.• It is the product of three decreasing functions (1/x, 1/ ln x, and 1/ ln ln x), somust itself be decreasing.Thus, using the integral test, we checkZ∞3dxx ln x ln(ln x)=Z∞ln ln 3duu= ∞where we substitute u = ln(ln x). Thus the original series also


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Berkeley MATH 1B - Math 1B Section 107 Quiz

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