1873_solutions

# Given that u is a differentiable function on an

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Unformatted text preview: 1 n x2  1 x 2j 3j x2 j1  3j 1 x 2j j1 2x 2  5x 4  8x 6    3n 2x 4  5x 6  8x 8    3n 1 x 2n  2x 2  3 x 4  x 6    x 2n 3n 3x 4 1 x 2n 1 x2 41 x 2n 3x 1 x 2 n 2 1 x 2 n 2  2x 2   2x 2   2x 2  3x 4 3x 2n4 and the final result follows by letting n 3n 3n 1 x 2n2 1 x 2 1 x2 3n 1 x 2n2 x 2n4 1 x2 Ý. Exercises on the Integral Test 1. For every positive integer n we have 2n jn1 2n 1 j j1 n 1 j log 2n j1 320 1 j log n  log 2. 1 x 2 n 2 Using this fact, deduce that 2n lim nÝ 1  log 2. j jn1 The identity 2n 2n jn1 1 j n 1 j j1 1 j log 2n j1 log n  log 2. log n  . is clear and the final result follows at once from the fact that 2n lim nÝ j1 n 1 j nÝ lim log 2n j1 1 j 2. Explain why 2n 1 j j1 2n j1  jn1 1 j for very positive integer n and deduce that 2n lim nÝ 1 j j1 j1  log 2. Deduce that 2 n 1 lim nÝ 1 j j1 j1 is also equal to log 2 and conclude that Ý 1 j j1 2n 1 j j1 j1 j1  log 2. n n 1  j1 2j 1 j1 n n 1  j1 2j 1 2n  j1 1 2j n 1 2j  j1 n 1 j j1 2 j1 1 2j 2n 1 j jn1 1 j The fact that 2n lim nÝ j1 1 j j1  log 2. follows at once from Exercise 1. Since 2 n 1 lim nÝ j1 1 j j1 2n nÝ lim j1 1 j j1 we conclude that 2 n 1 lim nÝ j1 Finally, to prove that 321 1 j j1  log 2.  1 2n  1 Ý j1 suppose that j1 1 j  log 2,  0. Choose N such that the inequalities 2n 1 j 1 j j1 1 j j1 j1 j1 log 2  and 2 n 1 j1 hold whenever n N. We see that n j1 whenever n log 2  log 2  2N  1. 3. Express the rational expression 1 j 2j 1 in partial fractions and show that if n is any positive integer we have n 2n 1 j 2j j1 2 1 jn1 1. j Deduce that Ý 1 j1 j 2j  log 4. 1 Solution: From the identity 1 j 2j we see that if n is any positive integer then n 2j j 2j n 2  1 1 j 1 n 1 j1 2  1 j1 2j 1 j1 n 1 j n 2  j1 2j 1 2n  j1  j1 n 2 2j n 2 j n j1 2 2j j1 2n j1 2 2 j jn1 1 j From Exercise 1 we deduce that Ý j1 2n 1 j 2j 1  n Ý2 lim jn1 1  2 log 2  log 4 j 4. Evaluate the sums Ý n1 Ý 1 n 2n  1 and n1 Solution: From the identity 322 1 . n 4n 2 1 1 j 1 j 1 j 2j  1 we see that if n is any positive integer then n n j1 1 j 2j  1 2 2j  1 n 1 j  j1 2 2j  1 j1 n n 1 j  j1 2 j j1 j1 2 n 1 n  n 2 2j j1 n 2 2j j1 j1 2 2j  1 2n 2 2  2 j jn1 1 j 2 2n  1 2 2 log 4. Therefore Ý j1 2n 1 j 2j  1 nÝ 2 lim 2 jn1 2 2n  1 1 j We conclude that Ý Ý 1 j 2j j1 1 1 j 2j  1 j1  4 log 2 2 and combining the terms on the left side we obtain Ý j1 5. Prove that if 1 q 1 j 4j 2 1  2 log 2 1. p then pn lim nÝ jqn1 1  log p . q j We see that pn lim nÝ pn jqn1 1  lim nÝ j qn 1 j j1 1 j j1 pn nÝ lim  j1 qn 1 j log pn j1 p   log q 1 j log qn p  log q p  log q . 6. Given that p  1, evaluate the limit np lim nÝ jn1 1. j log j Solution: We know that the limit n lim nÝ exists and is finite. Now whenever n np jn1 j2 1 j log j 2 we have np 1 j log j j2 log log n n 1 j log j log log n p j2 323 1 j log j log log n  log p and we co...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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