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quiz 5.2

# quiz 5.2 - note on p 699 for more details 2(3 pts Show that...

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Math 1B: Calculus March 3, 2010 Quiz 5 Lecturer: Prof. Mina Aganagic GSI: Gary Sivek Name: Answers 1. (2 pts) Is n =2 1 n (ln n ) 2 convergent or divergent? Justify your answer. Since x and ln x are increasing functions of x , it follows that 1 x (ln x ) 2 is a decreasing function of x whenever x > 1, where it is also continuous and positive. Then by the integral test, it converges if and only if the following integral converges: 2 dx x (ln x ) 2 = lim t →∞ t 2 dx x (ln x ) 2 = lim t →∞ ln t ln 2 du u 2 ( u = ln x, du = dx/x ) Since lim t →∞ ln t = , this last integral is simply ln 2 du/u 2 , which converges by the “ p -test” of § 7.8. Thus the series converges as well. Note that we have played fast and loose with the lower bounds for the index of summation and the region of integration here, but this is easily justified – see the
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Unformatted text preview: note on p. 699 for more details. 2. (3 pts) Show that the following series is divergent or Fnd the sum: ∞ ± n =2 ln n n + 1 Using our log rules, we Fnd a telescoping sum: s k = k ± n =2 ln n n + 1 = k ± n =2 (ln n-ln ( n + 1)) = k ± n =2 ln n-k ± n =2 ln ( n + 1) = k ± n =2 ln n-k +1 ± n =3 ln n s k = ln 2-ln ( k + 1) ±rom this we see that lim k →∞ s k =-∞ , so the sum diverges . This examples shows why we have to be careful to deFne convergence as a limit of partial sums this way. Otherwise, if we just blindly cancel out terms in an inFnite sum, we might claim that the series converges to ln 2, which is positive, even though every summand is negative!...
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