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Chapter_5

# Chapter_5 - V 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18...

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V 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 4 5 6 7 8 1

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V.1.1 1 2 3 P L K (Harmonic Series) Show that n X k =1 1 k ° log n: Deduce that the series P 1 k =1 1 k does not converge. Hint . Use the estimate 1 k ° Z k +1 k 1 x dx: Solution Let k be a nonnegative integer. Then since the function x 7! 1 =x is decreasing on the positive reals, we have for all x 2 [ k; k + 1] that 1 =x ± 1 =k . Hence locking at the oversum in the interval [ k; k + 1] 1 k (( k + 1) ² k ) = 1 k ° Z k +1 k 1 x dx: Therefore, for n ° 1 n X k =1 1 k ° n X k =1 Z k +1 k 1 x dx = Z n +1 1 1 x dx = log ( n + 1) ° log n: Where the last inequality follows from the fact that x 7²! log x is an increas- ing function on the positive reals. And thus is the sum P n k =1 1 k have not limit as n ! 1 and does not converge. 2
V.1.2 1 2 3 P L K Show that if p < 1 , then the series P 1 k =1 1 =k p diverges. Hint. Use Exercise 1 and the comparison test. Solution If k > 1 then, k p < k for 0 < p < 1 , then 1 k p > 1 k : Thus we have 1 X k =1 1 k p > 1 X k =1 1 k : By the comparison test, we have that the since the sum P n k =1 1 =k diverge as n ! 1 by Exercise 1, also the sum P n k =1 1 =k p diverge as n ! 1 , which was to be shown. 3

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V.1.3 1 2 3 P L K Show that if p > 1 , then the series P 1 k =1 1 =k p converges to S , where ° ° ° ° ° S ² n X k =1 1 k p ° ° ° ° ° < 1 ( p ² 1) n p ° 1 : Hint. Use the estimate 1 k p < R k k ° 1 dx x p . Solution Let k be a nonnegative integer. Then since the function x 7! 1 =x is decreasing on the positive reals, we have for all x 2 [ k ² 1 ; k ] that 1 =k ± 1 =x . Becauce p > 1 , it follows that 1 =k p ± 1 =x p . Hence locking att the undersum in the interval [ k ² 1 ; k ] 1 k p ( k ² ( k ² 1)) = 1 k p ± Z k k ° 1 1 x p dx Therefore n X k =1 1 k p ± n X k =1 Z k k ° 1 dx x p = 1+ Z n 1 dx x p = 1+ ± x ° p +1 ² p + 1 ² n 1 = 1+ 1 p ² 1 ² n ° p +1 p ² 1 ± 1+ 1 p ² 1 : Hence the serie converge to S , where S ± 1 + 1 p ² 1 = p p ² 1 ; since its terms are > 0 . Now we have that N X k = n +1 1 k p 6 Z N n dx x p = 1 + ± x ° p +1 ² p + 1 ² N n = n ° p +1 ² N ° p +1 p ² 1 ! n ° p +1 p ² 1 as N ! 1 . Thus we have 4
N X k = n +1 1 k p = S ² n X k =1 1 k p 6 n ° p +1 p ² 1 = 1 ( p ² 1) n p ° 1 thus ° ° ° ° ° S ² N X k = n +1 1 k p ° ° ° ° ° ± 1 ( p ² 1) n p ° 1 : 5

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V.1.4 1 2 3 P L K Show that the series 1 X k =1 ( ² 1) k +1 k = 1 ² 1 2 + 1 3 ² 1 4 + ³ ³ ³ converges. Hint. Show that the partial sums of the series satisfy S 2 < S 4 < S 6 < ³ ³ ³ < S 5 < S 3 < S 1 . Solution We de°ne the sum S n by S n = n X k =1 ( ² 1) k +1 k = 1 ² 1 2 + 1 3 ² 1 4 + : : : : This is the ±alternating series test², which is standard signs alternate, and terms ! 0 , j terms j # , so series converge. 6
V.1.5 1 2 3 P L K Show that the series 1 + 1 3 ² 1 2 + 1 5 + 1 7 ² 1 4 + 1 9 + 1 11 ² 1 6 + ³ ³ ³ converges to 3 S= 2 , where S is the sum of the series in Exercise 4. (It turns out that S = log 2 .) Hint. Organize the terms in the series in Exercise 4 in groups of four, and relate it to the groups of three in the above series.

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