MATH 105 Fall 2009 Midterm 2 Practice Exam Solutions - 4(1)n n 2(1)n = lim does not exist n n(n n 1\/n The sequence diverges(oscillates lim 18 n=1

MATH 105 Fall 2009 Midterm 2 Practice Exam Solutions -...

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Unformatted text preview: 4. lim n →∞ (-1 ) n n 2 π n ( n-π) = lim n →∞ (-1 ) n 1-(π/ n ) does not exist. The sequence diverges (oscillates). 18. ∞ summationdisplay n = 1 (-1 ) n 2 n-n converges absolutely by comparison with the convergent geometric series ∞ summationdisplay n = 1 1 2 n , because lim n →∞ vextendsingle vextendsingle vextendsingle vextendsingle (-1 ) n 2 n-n vextendsingle vextendsingle vextendsingle vextendsingle 1 2 n = lim n →∞ 1 1-n 2 n = 1 . 19. ∞ summationdisplay n = 1 (-1 ) n-1 ln ln n converges by the alternating series test, but the convergence is only conditional since ∞ summationdisplay n = 1 1 ln ln n diverges to infinity by comparison with the divergent harmonic series ∞ summationdisplay n = 1 1 n . (Note that ln ln n < n for all n ≥ 1.) 10. Since r 2 = 2 cos 2 θ meets r = 1 at θ = ± π 6 and ± 5 π 6 , the area inside the lemniscate and outside the circle is 4 × 1 2 integraldisplay π/ 6 bracketleftbig 2 cos 2 θ-1 2 bracketrightbig d θ = 2 sin 2...
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