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# 13 compute lim n z n 1 n log n x log n x 4 dx

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13. Compute lim n →∞ Z n 1 n [log( n + x ) - log n ] x 4 dx Solution. Notice that n [log( n + x ) - log n ] = log ‡‡ 1 + x n · n · log e x = x as n → ∞ . It is possible to prove, but it it isn’t as trivial as it looks, that the sequence '( 1 + x n ) n increases to e x , hence { n [log( n + x ) - log n ] } increases to x in [1 , ]. It is however an easy Calculus 1 exercise to prove that 1 + x n · n e x for all n N ∪ { 0 } , x 0. In fact, we proceed by induction on n proving that if g n ( x ) = e x - 1 + x n · n , 5

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then g n ( x ) 0 for all x 0. The case n = 0 is trivial; g 0 ( x ) = e x - 1 0 for all x 0. Assume the result proved up to some n 0. Consider g n +1 . We have g n +1 (0) = 0 and g 0 n +1 ( x ) = d dx ˆ e x - 1 + x n + 1 n +1 ! = e x - 1 + x n + 1 n e x - 1 + x n · n = g n ( x ) . The last in equality is due to 1 + x/ ( n + 1) 1 + x/n , thus (1 + x/ ( n + 1)) n (1 + x/n ) n . By the induction hypothesis, g 0 n +1 ( x ) 0 for all x 0, thus g n +1 increases in [0 , ); since g n +1 (0) = 0, we conclude g n +1 ( x ) 0 for all x 0. Having established this, we notice that Z n 1 n [log( n + x ) - log n ] x 4 dx = Z [1 , ) f n ( x ) dx, where f n ( x ) = χ [1 ,n ] ( x ) n [log( n + x ) - log n ] x 4 . For every x [1 , ] we have χ [1 ,n ] ( x ) = 1 once n > x , thus lim n →∞ f n ( x ) = lim n →∞ n [log( n + x ) - log n ] x 4 = x x 4 = 1 x 3 . Moreover | f n ( x ) | ≤ χ [1 ,n ] ( x ) log e x x 4 log e x x 4 = 1 x 3 for all n N , x 1. Because x 7→ 1 /x 3 is integrable, Lebesgue’s dominated convergence theorem applies to give lim n →∞ Z n 1 n [log( n + x ) - log n ] x 4 dx = lim n →∞ Z 1 χ [1 ,n ] ( x ) n [log( n + x ) - log n ] x 4 dx = Z 1 lim n →∞ χ [1 ,n ] ( x ) n [log( n + x ) - log n ] x 4 dx = Z 1 1 x 3 dx = 1 2 . Fo full credit I needed to see some proof that either dominated convergence or monotone convergence applies. 14. Recall that if f : [ a, b ] R , the total variation of f on [ a, b ] is defined by V b a ( f ) = sup { n X i =1 | f ( x i ) - f ( x i - 1 ) | : x 0 = a < x 1 < . . . < x n = b, n N } . The function f is said to be of bounded variation if V b a ( f ) < . Prove: If f : [ a, b ] R is continuous, if f is differentiable at all points of ( a, b ) and f 0 is a bounded continuous function on ( a, b ), then V b a ( f ) = Z b a | f 0 ( x ) | dx. Solution. To show V b a ( f ) int b a | f 0 | is easy. The converse inequality is hard. Maybe quite hard. Let P = { x 0 = a < x 1 < · · · < x n = b } be a partition of [ a, b ]. because f is everywhere differentiable and f 0 is continuous, we have n X i =1 | f ( x i ) - f ( x i - 1 ) | = n X i =1 fl fl fl fl fl Z x i x i - 1 f 0 ( x ) dx fl fl fl fl fl n X i =1 Z x i x i - 1 | f 0 ( x ) | dx = Z b a | f 0 ( x ) | dx. Since the partition was arbitrary, we see that R b a | f 0 | dx is an upper bound of the set of which V b a ( f ) is the supremum. This proves (the easy part) V b a ( f ) Z b a | f 0 ( x ) | dx. 6
For the converse inequality, let A = { x ( a, b ) : f 0 ( x ) > 0 } , B = { x ( a, b ) f 0 ( x ) < 0 } . Because f 0 is continuous, the sets A, B are open, thus they can be expressed as countable unions of pairwise disjoint open intervals. Say, A = S n =1 I n , B = n =1 J n where I n = ( a n , b n ), J n = ( c n , d n ) for n N ; I n I k = , J n J k = if n 6 = k . Also, since A B = , I n J k = for all k, n (even n = k ).

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• Spring '11
• Speinklo
• Empty set, Open set, Topological space, Dominated convergence theorem, Lebesgue integration, Non-measurable set

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