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Unformatted text preview: 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 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) 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 7! 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 < 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 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 & n X k =1 1 k p & & & & & < 1 ( p & 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 & 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 & 1 ;k ] 1 k p ( k & ( k & 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 & p + 1 n 1 = 1+ 1 p & 1 & n & p +1 p & 1 1+ 1 p & 1 : Hence the serie converge to S , where S 1 + 1 p & 1 = p p & 1 ; since its terms are > . Now we have that N X k = n +1 1 k p 6 Z N n dx x p = 1 + x & p +1 & p + 1 N n = n & p +1 & N & p +1 p & 1 ! n & p +1 p & 1 as N ! 1 . Thus we have 4 N X k = n +1 1 k p = S & n X k =1 1 k p 6 n & p +1 p & 1 = 1 ( p & 1) n p & 1 thus & & & & & S & N X k = n +1 1 k p & & & & & 1 ( p & 1) n p & 1 : 5 V.1.4 1 2 3 P L K Show that the series 1 X k =1 ( & 1) k +1 k = 1 & 1 2 + 1 3 & 1 4 + converges. Hint. Show that the partial sums of the series satisfy S 2 < S 4 < S 6 < < S 5 < S 3 < S 1 . Solution We de&ne the sum S n by S n = n X k =1 ( & 1) k +1 k = 1 & 1 2 + 1 3 & 1 4 + :::: This is the alternating series test, which is standard signs alternate, and terms ! , j terms j # , so series converge....
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This note was uploaded on 11/13/2010 for the course MATH 132 taught by Professor Grossman during the Spring '08 term at UCLA.
 Spring '08
 Grossman
 Harmonic Series

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