# sol06 - Math 104, Solution to Homework 6 Instructor:...

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Unformatted text preview: Math 104, Solution to Homework 6 Instructor: Guoliang Wu July 15, 2009 Ross, K. A., Elementary Analysis: The theory of calculus : 12.12 Proof. (a) If n &gt; M &gt; N &gt; , then n = 1 n ( s 1 + s 2 + + s N + s N +1 + + s n ) = 1 n ( s 1 + s 2 + + s N ) + 1 n ( s N +1 + + s n ) 1 M ( s 1 + s 2 + + s N ) + 1 n ( n- N ) sup { s n : n &gt; N } 1 M ( s 1 + s 2 + + s N ) + sup { s n : n &gt; N } Therefore sup { n : n &gt; M } 1 M ( s 1 + s 2 + + s N ) + sup { s n : n &gt; N } . Note that sup { n : n &gt; M } is decreasing in M , we deduce that lim sup n 1 M ( s 1 + s 2 + + s N ) + sup { s n : n &gt; N } , for any M &gt; N &gt; . Fix any N &gt; . We regard the right hand side as a sequence in M and take the limit as M , we have lim sup n sup { s n : n &gt; N } . Finally, regard the right hand side as a sequence in N and take the limit as N , we obtain lim sup n lim sup s n . The second inequality lim inf n lim sup n is obvious. 1 The first inequality can be derived from the third one. Let s n =- s n and n =- n for all n , then n = 1 n ( s 1 + s 2 + s n ) ....
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## sol06 - Math 104, Solution to Homework 6 Instructor:...

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