# M 408D HW 2 Solutions - cruz(fmc326 HW02 kalahurka(55295...

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cruz (fmc326) – HW02 – kalahurka – (55295) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Which, if any, of the following series converge? ( A ) summationdisplay m =2 5 ln m m 2 ( B ) summationdisplay n =2 1 n (ln n ) 2 1. neither A nor B 2. A and B correct 3. A only 4. B only Explanation: ( A ) The function f ( x ) = 5 ln x x 2 is continous and positive on [2 , ); in addi- tion, since f ( x ) = 5 parenleftbigg 1 2 ln x x 3 parenrightbigg < 0 on [2 , ), f is also decreasing on this inter- val. This suggests applying the Integral Test. Now, after Integration by Parts, we see that integraldisplay t 2 f ( x ) dx = 5 bracketleftBig ln x x 1 x bracketrightBig t 2 , and so integraldisplay 2 f ( x ) dx = 5(1 + ln 2) . The Integral Test thus ensures that the given series converges . ( B ) The given series is of the form summationdisplay n =2 f ( x ) with f ( x ) = 1 x (ln x ) 2 , x 2 . Since f is positive and decreasing on [2 , ), Integral Test ensures that the given series is convergent if the improper integral I = integraldisplay 2 f ( x ) dx converges. But, after the substitution u = ln x we see that I = lim t → ∞ bracketleftBig 1 u bracketrightBig t ln2 = 1 ln 2 . Consequently, the series converges . keywords: 002 10.0points What is the smallest number of terms of the series summationdisplay m =1 2 ( m + 1)(ln( m + 1)) 2 you would need to add to find its sum to within 1 10 ? 1. e 20 terms correct 2. e 40 terms 3. e 35 terms 4. e 30 terms
cruz (fmc326) – HW02 – kalahurka – (55295) 2 5. e 25 terms Explanation: If f is a function which is continuous, pos- itive and decreasing on [1 , ), then the in- equalities summationdisplay m = M +1 f ( m ) < integraldisplay M f ( x ) dx < summationdisplay m = M f ( m ) hold for any M 1. Thus, if s = summationdisplay m =1 f ( m ) , S M = M summationdisplay m =1 f ( m ) , then s S M < integraldisplay M f ( m ) dx < s S M 1 . Consequently, if we want the smallest value of M so that S M is within 1 10 of the sum s of the series, we need to find the smallest value of M so that integraldisplay M f ( x ) dx 1 10 . But for the given series f ( x ) = 2 ( x + 1)(ln( x + 1)) 2 , a change of variable u = ln( x +1) ensures that integraldisplay M f ( x ) dx = integraldisplay M , 2 ( x + 1)(ln( x + 1)) 2 dx = lim t → ∞ bracketleftBig 2 u bracketrightBig t ln( M +1) = 2 ln( M + 1) . Thus integraldisplay M f ( x ) dx 1 10 = 2 ln( M + 1) 1 10 , and so the smallest number of terms you would need to add is e 20 terms . 003 10.0points Use the Remainder Estimate integraldisplay n +1 f ( x ) dx R n integraldisplay n f ( x ) dx for the Integral Test applied with f ( x ) = 4 x 3 to determine the smallest value of n so that the error R n = s S n in using the n th partial sum S n to estimate the sum s of the series summationdisplay k =1 4 k 3 is less than 1 / 10 2 . 1. n = 13 2. n = 14 3. n = 15 correct 4. n = 11 5. n = 12 Explanation: We have to find the smallest value of n so that s S n integraldisplay n 4 x 3 dx < 1 10 2 . Now integraldisplay n 4 x 3 dx = lim t → ∞ integraldisplay t n 4 x 3 dx , while integraldisplay t n 4 x 3 dx = bracketleftBig 2 x 2 bracketrightBig t n = 2 n 2 2 t 2 −→ 2 n 2 as t → ∞ . Thus we have to solve the inequal- ity 2 n 2 < 1 10 2 .
cruz (fmc326) – HW02 – kalahurka – (55295) 3