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Math 321 - The Dirichlet Test

Math 321 - The Dirichlet Test - The Dirichlet Test...

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The Dirichlet Test Theorem (The Dirichlet Test) Let X be a metric space. If the functions f n : X C , g n : X IR , n IN obey F n ( x ) = n m =1 f m ( x ) is bounded uniformly in n and x g n +1 ( x ) g n ( x ) for all x X and n b g n ( x ) B n converges uniformly to zero on X then n =1 f n ( x ) g n ( x ) converges uniformly on X . Proof: The trick to this proof is the summation by parts formula, which we now derive. s n ( x ) = n s k =1 f k ( x ) g k ( x ) = F 1 ( x ) g 1 ( x ) + n s k =2 [ F k ( x ) F k - 1 ( x )] g k ( x ) = F 1 ( x ) g 1 ( x ) + n s k =2 F k ( x ) g k ( x ) n s k =2 F k - 1 ( x ) g k ( x ) = n s k =1 F k ( x ) g k ( x ) n - 1 s k =1 F k ( x ) g k +1 ( x ) = n s k =1 F k ( x ) ± g k ( x ) g k +1 ( x ) ² + g n +1 ( x ) F n ( x ) So if m > n , the diFerence between the m th and n th partial sums is s m ( x ) s n ( x ) = m s k = n +1 F k ( x ) ± g k ( x ) g k +1 ( x ) ² + g m +1 ( x ) F m ( x ) g n +1 ( x ) F n ( x ) If M = sup b | F n ( x ) | v v x X, n B , v v s m ( x ) s n ( x ) v v M m s k = n +1 ± g k ( x ) g k +1 ( x ) ² + Mg m +1 ( x ) + n +1 ( x ) = M ± g n +1 ( x ) g m +1 ( x ) ² + m +1 ( x ) + n +1 ( x ) = 2 n +1 ( x ) (1) since g m +1 ( x ) 0, g n +1 ( x )
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Math 321 - The Dirichlet Test - The Dirichlet Test...

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