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exam2f09sol

# exam2f09sol - MAT 127 Calculus C Fall 2009 Solutions to...

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MAT 127: Calculus C, Fall 2009 Solutions to Midterm II Problem 1 (15pts) Determine whether each of the following sequences converges or diverges; justify your answer. You do not have to determine the limit if the sequence converges. (a; 5pts) a n = ( - 1) n n 4 3 n 4 + 1 diverges because a n = ( - 1) n n 4 3 n 4 + 1 = ( - 1) n n 4 /n 4 3 n 4 /n 4 + 1 /n 4 = ( - 1) n 1 3 + 1 /n 4 ; as n -→ ∞ the fraction approaches 1 / 3, while the entire expression keeps on jumping from near 1 / 3 (even n ) to near - 1 / 3 (odd n ). Grading: correct answer 2pts; dividing by n 4 , obtaining 1/3 as limit of the fraction, and some comment on the sign 1pt each (-1pt for mce ); wrong answer with complete explanation containing mce in division by n 4 or forgetting ( - 1) n max 2pts. mce = mi- nor computa- tional error (b; 5pts) a n = 1 + sin( nπ/ 2) ln n n converges because (ln n ) /n approaches 0, while | sin( nπ/ 2) | ≤ 1; so sin( nπ/ 2)(ln n ) /n -→ 0 and a n -→ 1. Grading: correct answer 2pts; correct statements regarding (ln n ) /n and sin( nπ/ 2) 2pts and 1pt, respectively; so . . . not required; wrong answer with complete explanation misidentifying lim n -→∞ (ln n ) /n max 2pts. (c; 5pts) a 1 = 1 , a n +1 = a n + 9 2 converges because a n 9 for all n : this is true for n =1; if this is true for n , a n +1 = ( a n +9) / 2 (9+9) / 2 = 9 and so this is true for n +1; a n a n +1 : a n +1 = ( a n +9) / 2 ( a n + a n ) / 2 = a n . So { a n } converges by the Monotonic Sequence Theorem. Grading: correct answer 2pts; bounded above 2pts; increasing 1pt; last sentence not required; wrong answer 0pts regardless of explanation. Alternatively (bonus 2pts, whether or not in addition to a standard argument): the sequence { a n } is obtained by starting with some a 1 < 9, then taking a 2 to be the mid-point of the segment [ a 1 , 9], a 3 to be the mid-point of the segment [ a 2 , 9], and so on. This sequence of consecutive mid-points approaches 9, since at each step the new distance to 9 is reduced to 1/2 of the old distance to 9. Grading: the substance must be identical to the above for full credit, but could be worded (with a bit of care) in terms of the sequence b n =9 - a n ; 2nd half of last sentence not required.

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Problem 2 (20pts) Each of the following sequences converges (you do not need to check this); determine its limit (show your work!). (a; 6pts) a n = n 1 + 4 n 2 + 1 a n = n/n 1 /n + 4 n 2 + 1 /n = 1 1 /n + radicalbig 4 n 2 /n 2 + 1 /n 2 = 1 1 /n + radicalbig 4 + 1 /n 2 So a n -→ 1 / (0 + 4 + 0) = 1 / 2 Grading: correct answer 3pts; dividing by n 1pt; 2pts to the answer; mce -1pt (from 6pts); taking n under without squaring it resulting in limit of 0 -2pts (from 6pts) mce = mi- nor computa- tional error (b; 7pts) a n = 5 n n ! (reminder: n ! = 1 · 2 · . . . · n ) 1st Approach. Since the limit exists and a n +1 = 5 n + 1 a n , ok without n -→∞ lim n -→∞ a n = lim n -→∞ a n +1 = lim n -→∞ parenleftbigg 5 n + 1 a n parenrightbigg = parenleftbigg lim n -→∞ 5 n + 1 parenrightbigg · ( lim n -→∞ a n ) = 0 · ( lim n -→∞ a n ) = 0 2nd Approach. a n 0 for all n and a 5+ n = 5 5 5!
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