Description: Lemma for iseralt . From iseraltlem2 , we have ( -u 1 ^ n ) x. S ( n + 2 k ) <_ ( -u 1 ^ n ) x. S ( n ) and ( -u 1 ^ n ) x. S ( n + 1 ) <_ ( -u 1 ^ n ) x. S ( n + 2 k + 1 ) , and we also have ( -u 1 ^ n ) x. S ( n + 1 ) = ( -u 1 ^ n ) x. S ( n ) - G ( n + 1 ) for each n by the definition of the partial sum S , so combining the inequalities we get ( -u 1 ^ n ) x. S ( n ) - G ( n + 1 ) = ( -u 1 ^ n ) x. S ( n + 1 ) <_ ( -u 1 ^ n ) x. S ( n + 2 k + 1 ) = ( -u 1 ^ n ) x. S ( n + 2 k ) - G ( n + 2 k + 1 ) <_ ( -u 1 ^ n ) x. S ( n + 2 k ) <_ ( -u 1 ^ n ) x. S ( n ) <_ ( -u 1 ^ n ) x. S ( n ) + G ( n + 1 ) , so | ( -u 1 ^ n ) x. S ( n + 2 k + 1 ) - ( -u 1 ^ n ) x. S ( n ) | = | S ( n + 2 k + 1 ) - S ( n ) | <_ G ( n + 1 ) and | ( -u 1 ^ n ) x. S ( n + 2 k ) - ( -u 1 ^ n ) x. S ( n ) | = | S ( n + 2 k ) - S ( n ) | <_ G ( n + 1 ) . Thus, both even and odd partial sums are Cauchy if G converges to 0 . (Contributed by Mario Carneiro, 6-Apr-2015)
Ref | Expression | ||
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Hypotheses | iseralt.1 | |
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iseralt.2 | |
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iseralt.3 | |
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iseralt.4 | |
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iseralt.5 | |
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iseralt.6 | |
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Assertion | iseraltlem3 | |